To solve this, we first express the problem in a new way: Instead of having a laser bouncing between two mirrors, we have a laser moving straight through a different space; to accomplish this, we draw what you would see if you were standing in the setup - that is, where the mirrors are, we place a new, reflected image of the whole space! That is, we make copies of the $1^{\circ}$ wedge by mirroring it over the rays that bound it. The result looks something like this:
where the line segments represent the mirrors and the red line is the laser. This is basically what you'd see if you were standing inside such a setup (although you'd see many copies of yourself too). We may regard this as a sort of "covering space" for the old one as each point in the old one is copied many times into the new one.
This makes the problem simple: How many consecutive line segments can the red line cross? Well, let's call the distance from the center of the circle to the laser $d$. And, for simplicitly, lets call the radius of the inner circle $1$ and the outer circle $2$, since we only need to preserve the ratio between those measurements (which is 10m : 20m originally). Then, a laser crosses a line segment measuring an angle of $\theta$ to it if:
$$\sin(\theta)\leq d\leq 2\sin(\theta)$$
We thus require that the smallest value of $2\sin(\theta)$ of any line we cross is at least equal to than the greatest value of $\sin(\theta)$ in the lines that we cross - otherwise, no suitable $d$ exists.
Using this inequality, suppose we have that the leftmost line $\alpha$ to the line, and the rightmost line is at an angle of $\beta$ (which will be greater than $\alpha$). We may assume that $|\alpha|\geq|\beta|$, as the two are related by reflection. Moreover, we may then write $\alpha=(n+c)^{\circ}$ where $n$ is an integer and $-\frac{1}2\leq c\leq \frac{1}2$ - meaning $90^{\circ}+c$ will be the maximum $\sin(\theta)$ to appear. Then, all we need to satisfy for the intervals for $d$ to intersect is:
$$\sin(90+c)\leq 2\sin(\alpha)$$
$$\sin(90+c)\leq 2\sin(\beta)$$
Now, I'm sure there's an elegant way to solve this. I'm kind of a hack though, so we'll just make some guesses: Suppose that $c$ were $\frac{1}2$. Then we can find that the integer $n$ satisfying
$$\sin(90+\frac{1}2)\leq 2\sin(90+n+\frac{1}2)$$
are exactly those in the interval $[-60,59]$, meaning we can, from any point, cross $60$ to either side of it (as $c=\frac{1}2$ is the most extreme value possible) - for a maximum of $121$ reflections.
However, letting $c$ be zero, it's a fairly common fact from highschool trigonometry that $\sin(30^{\circ})=\sin(150^{\circ})=\frac{1}2$ that $\alpha=30^{\circ}$ meaning that we can hit every mirror making an angle between $30^{\circ}$ and $150^{\circ}$ with the laser - this achieves the maximum of $121$ reflections. To do this, we do as follows:
Aim the laser at the outer edge of one of the mirrors, impacting the mirror at a $30^{\circ}$ angle. It will reflect $121$ times.
A picture of this would probably just look like a lot of red lines. However, the same logic applies if the angle between the mirrors were $5^{\circ}$ and predicts $25$ reflections. That looks like this:
Notice that it hits one mirror perpendicularly, and hence retraces its steps. In other words, some advice:
Don't look in the direction you point the laser.
