Answer
${\bf 18\;000}\;\rm photon$
Work Step by Step
We know that the probability of finding a photon within a narrow region of width $\delta x$ at position $x$ is given by
$$\text{Prob}( \text{in } \delta x \text{ at } x) \propto |A(x)|^2 \delta x$$
This means that the probability of finding a photon in a small region $\delta x $ at position $x $ is proportional to the square of the amplitude of the light wave $ |A(x)|^2 $ times the width of the region $\delta x $.
So the ratio of probabilities at two different positions $x_1$ and $x_2$:
$$ \frac{\text{Prob(in } \delta x_1 \text{ at } x_1)}{\text{Prob(in } \delta x_2 \text{ at } x_2)} = \frac{|A(x_1)|^2 \delta x}{|A(x_2)|^2 \delta x} $$
Since the width $ \delta x $ is the same, they cancel out,
$$ \frac{\text{Prob(in } \delta x_1 \text{ at } x_1)}{\text{Prob(in } \delta x_2 \text{ at } x_2)} = \frac{|A(x_1)|^2}{|A(x_2)|^2} $$
Hence,
$$\dfrac{\dfrac{N_1}{N_{tot}}}{\dfrac{N_2}{N_{tot}}} = \frac{|A(x_1)|^2}{|A(x_2)|^2} $$
where $N_{tot}$ is the total number of photons, $ N_1$ is the number of photons detected at position $ x_1 $ in the region $\delta x$, and $ N_2 $ is the number of photons detected at position $ x_2 $ in the region $\delta x$
$N_{tot}$ they cancel out,
$$\dfrac{N_1}{ N_2} = \frac{|A(x_1)|^2}{|A(x_2)|^2}$$
Thus,
$$N_2=\dfrac{N_1|A(x_2)|^2}{|A(x_1)|^2}$$
Plug the known;
$$N_2=\dfrac{2000 (30)^2}{(10)^2}$$
$$N_2=\color{red}{\bf 18\;000}\;\rm photon$$
