Answer
$1.39\times 10^{-3}$
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 corral of edge length $L_x=L_y=L=150\;pm\;$. The energy state of the electron is $E_{1,3}$ $(n_x=1, n_y=3)$. The electron is detected by a probe of area, $dx dy=5\;pm\times 5\;pm$. 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, $dx dy=5\;pm\times 5\;pm$ centered at $xy$ coordinates $(0.200L, 0.800L)$ is
$p(x,y)={\frac{2}{150}}\sin^2(0.2\pi)\times{\frac{2}{150}}\sin^2(0.8\times3\pi)\times 25$
or, $p(x,y)\approx 1.39\times 10^{-3}$
