Answer
${\bf 100}\;\rm V/m$
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 $x_2$:
$$ \frac{\text{Prob(in } \delta x_1 \text{ at } x_1)}{\text{Prob(in } \delta x_2 \text{ at } x_2)} = \frac{|A(x_1)|^2 \delta x_1}{|A(x_2)|^2 \delta x_2} $$
Hence,
$$\dfrac{\dfrac{N_1}{N_{tot}}}{\dfrac{N_2}{N_{tot}}} = \frac{|A(x_1)|^2 \delta x_1}{|A(x_2)|^2 \delta x_2} $$
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,
$$\dfrac{N_1}{ N_2} =\frac{|A(x_1)|^2 \delta x_1}{|A(x_2)|^2 \delta x_2} $$
Solving for $|A(x_2)|$;
$$|A(x_2)|^2 =\dfrac{ N_2}{N_1} \frac{|A(x_1)|^2 \delta x_1}{ \delta x_2} $$
$$|A(x_1)| =\sqrt{\dfrac{ N_2}{N_1} \frac{|A(x_1)|^2 \delta x_1}{ \delta x_2} }$$
Plug the known;
$$|A(x_1)| =\sqrt{\dfrac{ 3000}{6000} \frac{(200)^2 (0.1)}{(0.2)} }=\color{red}{\bf 100}\;\rm V/m$$
