Answer
$27.61\;pm$
Work Step by Step
The normalized wavefunction $\psi_{nx,ny}$ for an electron in an infinite, two-dimensional potential well are given by
$\psi_{nx,ny}=\sqrt {\frac{2}{L_x}}\sin\Big(\frac{n_x\pi}{L_x}x\Big)\sqrt {\frac{2}{L_y}}\sin\Big(\frac{n_y\pi}{L_y}y\Big)$
where $n_x$ is a quantum number for well width $L_x$ and $n_y$ is a quantum number for well width $L_y$.
Now, the probability that the electron will be detected by a probe having the area of $dx dy$ is given by
$p(x,y)=\psi^2_{nx,ny} dx dy$
or, $p(x,y)={\frac{2}{L_x}}\sin^2\Big(\frac{n_x\pi}{L_x}x\Big){\frac{2}{L_y}}\sin^2\Big(\frac{n_y\pi}{L_y}y\Big)dxdy$
In our case, the electron trapped in the two-dimensional, infinite square potential well of edge length $L_x=L_y=L$. The electron is in the ground $(n_x=1, n_y=1)$. The electron is detected by a probe of area, $dx dy=400\;pm^2$. The area $dx dy$ is so narrow that we can take the probability density to be constant within it.
Therefore, the probability of finding electron within a area, $dx dy=400\;pm^2$ centered at $xy$ coordinates $(\frac{L}{8},\frac{L}{8})$ is
$p(x,y)={\frac{2}{L}}\sin^2(\frac{\pi}{8})\times{\frac{2}{L}}\sin^2(\frac{\pi}{8})\times 400$
or, $p(x,y)=\frac{1600}{L^2}\sin^4(\frac{\pi}{8})$
Now,
$p(x,y)=4.5\times 10^{-8}$
or, $\frac{1600}{L^2}\sin^4(\frac{\pi}{8})=4.5\times 10^{-8}$
or, $L=\sqrt {\frac{1600}{4.5\times 10^{-8}}\sin^4(\frac{\pi}{8})}\;pm$
or, $L=27.61\;pm$
