Answer
a) $\bf 3330$
b) $\bf 1110$
Work Step by Step
We know that the probability density is given by
$$\text{Prob}( \text{in } \delta x \text{ at } x)=P(x)\delta x=\dfrac{N }{N_{tot}}$$
where $P(x)=|\psi (x)|^2$, $N_{tot}$ is the total number of photons, and $N$ is the number of photons detected at position $\delta x$.
Hence,
$$|\psi (x)|^2 \delta x=\dfrac{N }{N_{tot}}$$
Solving for $N$ since the author asks about the expected number of photons that will land in some wide strip.
$$N=N_{tot}|\psi (x)|^2 \delta x\tag 1$$
$$\color{blue}{\bf [a]}$$
Now we need to find the expected number of photons that will land in a 0.010-mm wide strip at $x=0$ mm.
Plug the known into (1) where we found from the given graph that $|\psi (0)|^2=0.333$ mm$^{-1}$ at $x=0$
$$N=(1\times 10^6)(0.333)(0.01)$$
$$N=\color{red}{\bf 3330}\;\rm photon$$
$$\color{blue}{\bf [b]}$$
Plug the known into (1) where we found from the given graph that $\bullet|\psi (2)|^2\approx 0.333-\dfrac{0.333}{3}\times 2=0.111$ mm$^{-1}$ at $x=2$ mm
$$N=(1\times 10^6)(0.111)(0.01)$$
$$N=\color{red}{\bf 1110}\;\rm photon$$
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We use the straight line formula $y=mx+b$ and the slope of the line to find $|\psi (2)|^2$ at $x=2$.
