Observations to give lower and upper bounds:

1. $\lfloor x\rfloor\leq x$, so we must have $2020=x \left\lfloor x \left\lfloor x \left\lfloor x \right\rfloor \right\rfloor \right\rfloor\leq x^4$, therefore $x\geq\sqrt[4]{2020}=6.704\dots$

2. If $x\geq7$, then $x\lfloor x\rfloor\geq49$ and so on until $2020=x \left\lfloor x \left\lfloor x \left\lfloor x \right\rfloor \right\rfloor \right\rfloor\geq 7^4=2401$. Contradiction.

So we know for sure

$x$ is six point something and $\lfloor x\rfloor=6$. Also $6.704\dots\leq x<7$ means $40.224\dots\leq6x<42$, so $\lfloor x\lfloor x\rfloor\rfloor$ must be either $40$ or $41$.

Now the whole thing becomes

$2020=x \left\lfloor x (40\text{ or }41) \right\rfloor$. The thing inside this final floor sign is at least $40\times6.704\dots=268.162\dots$ and at most $41\times7=287$. Which means $x$ must be at least $2020\div287=7.038.$

Contradiction ... and now I realise my implicit assumption that

$x\geq0$.

---

Going back to those two observations at the beginning,

with the knowledge that $x$ is negative, we have $$x\geq-6\Rightarrow\lfloor x\rfloor\geq-6\Rightarrow x\lfloor x\rfloor\leq36\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor\geq-216\Rightarrow x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\leq 1296,$$ $$x\leq-7\Rightarrow\lfloor x\rfloor\leq-7\Rightarrow x\lfloor x\rfloor\geq49\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor\leq-343\Rightarrow x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\geq 2401,$$ $$\lfloor x\rfloor\leq x\Rightarrow x\lfloor x\rfloor\geq x^2\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor\leq x\lfloor x^2\rfloor\leq x^3\Rightarrow 2020=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\geq x^4,$$ so $-6.704\dots\leq x<-6$ and $\lfloor x\rfloor=-7$.

That means $42<x\lfloor x\rfloor\leq46.928$ and $43\leq\lfloor x\lfloor x\rfloor\rfloor\leq46$.

That means $-308.38\dots\leq x\lfloor x\lfloor x\rfloor\rfloor<-258$.

So we seek a number which,

when multiplied by an integer between $258$ and $308$, gives $2020$. Dividing $2020$ by $6$ and $7$ gives that this integer must be between $289$ and $336$. Going the other way, the bound of $308$ means $x\geq-\frac{2020}{308}=-6.558\dots$. Since this bound came from taking the fourth root, we expect $x$ should be close to it.

So we try just a few nearby values of the integer:

$x=-\frac{2020}{308}=-6.558\dots\Rightarrow x\lfloor x\rfloor=7\times6.558\dots=45.909\dots\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-45\times6.558\dots=295.13\dots$, too small.

$x=-\frac{2020}{307}=-6.580\dots\Rightarrow x\lfloor x\rfloor=7\times6.580\dots=46.059\dots\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-46\times6.580\dots=302.67\dots$, too small but much closer!

$x=-\frac{2020}{306}=-6.601\dots\Rightarrow x\lfloor x\rfloor=7\times6.601\dots=46.209\dots\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-46\times6.601\dots=303.66\dots$, too small but much closer!

$x=-\frac{2020}{305}=-6.623\dots\Rightarrow x\lfloor x\rfloor=7\times6.623\dots=46.361\dots\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-46\times6.623\dots=304.66\dots$, exactly right!

And we have the solution,

$x=-\frac{2020}{305}=-6.623\dots$

Observations to give lower and upper bounds:

1. $\lfloor x\rfloor\leq x$, so we must have $2020=x \left\lfloor x \left\lfloor x \left\lfloor x \right\rfloor \right\rfloor \right\rfloor\leq x^4$, therefore $x\geq\sqrt[4]{2020}=6.704$

2. If $x\geq7$, then $x\lfloor x\rfloor\geq49$ and so on until $2020=x \left\lfloor x \left\lfloor x \left\lfloor x \right\rfloor \right\rfloor \right\rfloor\geq 7^4=2401$. Contradiction.

So we know for sure

$x$ is six point something and $\lfloor x\rfloor=6$. Also $6.704\leq x<7$ means $40.224\leq6x<42$, so $\lfloor x\lfloor x\rfloor\rfloor$ must be either $40$ or $41$.

Now the whole thing becomes

$2020=x \left\lfloor x (40\text{ or }41) \right\rfloor$. The thing inside this final floor sign is at least $40\times6.704=268.16$ and at most $41\times7=287$. Which means $x$ must be at least $2020\div287=7.038$.

Contradiction ... and now I realise my implicit assumption that

$x\geq0$.

---

Going back to those two observations at the beginning,

with the knowledge that $x$ is negative, we have $$x\geq-6\Rightarrow\lfloor x\rfloor\geq-6\Rightarrow x\lfloor x\rfloor\leq36\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor\geq-216\Rightarrow x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\leq 1296,$$ $$x\leq-7\Rightarrow\lfloor x\rfloor\leq-7\Rightarrow x\lfloor x\rfloor\geq49\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor\leq-343\Rightarrow x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\geq 2401,$$ $$\lfloor x\rfloor\leq x\Rightarrow x\lfloor x\rfloor\geq x^2\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor\leq x\lfloor x^2\rfloor\leq x^3\Rightarrow 2020=x\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\geq x^4,$$ so $-6.704\leq x<-6$ and $\lfloor x\rfloor=-7$.

That means $42<x\lfloor x\rfloor\leq46.928$ and $43\leq\lfloor x\lfloor x\rfloor\rfloor\leq46$.

That means $-308.38\leq x\lfloor x\lfloor x\rfloor\rfloor<-258$.

So we seek a number which,

when multiplied by an integer between $258$ and $308$, gives $2020$. Dividing $2020$ by $6$ and $7$ gives that this integer must be between $289$ and $336$. Going the other way, the bound of $308$ means $x\geq-\frac{2020}{308}=-6.558$. Since this bound came from taking the fourth root, we expect $x$ should be close to it.

So we try just a few nearby values of the integer:

$x=-\frac{2020}{308}=-6.558\Rightarrow x\lfloor x\rfloor=7\times6.558=45.909\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-45\times6.558=295.13$, too small.

$x=-\frac{2020}{307}=-6.580\Rightarrow x\lfloor x\rfloor=7\times6.580=46.059\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-46\times6.580=302.67$, too small but much closer!

$x=-\frac{2020}{306}=-6.601\Rightarrow x\lfloor x\rfloor=7\times6.601=46.209\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-46\times6.601=303.66$, too small.

$x=-\frac{2020}{305}=-6.623\Rightarrow x\lfloor x\rfloor=7\times6.623=46.361\Rightarrow x\lfloor x\lfloor x\rfloor\rfloor=-46\times6.623=304.66$, exactly right!

And we have the solution,

$x=-\frac{2020}{305}=-6.623\dots$