Let's calculate the radius of the ring of 18 coins, taking the radius of the coin as $1$.
The centres of the coins form a regular 18-gon. Two adjacent coin centres and the origin form an isosceles triangle with an angle at the origin of $360/18=20$ and base length $1$. We then get $$ r \sin \frac{20}2 = \frac12$$
So $r=2.879$. This is the distance from the origin to the centres of the outer ring of coins. Note that this radius was rounded down so that the area to be filled will be underestimated.
Now let's try to fill the inside area by
dividing it into rings or annuli of width 1, and filling those with coins. So one ring has its coins at distance $1.879$, and inside that is another ring with the coins at distance $0.879$ from the origin.
We now use the same formula as before but calculate the angle from the radius instead of the radius from the angle.
$$ 1.879 \sin \frac{\alpha}{2} = \frac12\\ \alpha=30.9$$
This means we can fit $\lfloor \frac{360}{30.9} \rfloor = 11$ coins in the ring.
Similarly for the inner ring we get
$$ 0.879 \sin \frac{\alpha}{2} = \frac12\\ \alpha=69.3$$
This means we can fit $\lfloor \frac{360}{69.3} \rfloor = 5$ coins in the inner ring.
So it is fairly straightforward to fit 16 coins inside the ring of 18 coins.