Answer
$0.136\;eV$
Work Step by Step
The quantized energies for an electron trapped in a two dimensional infinite potential well are
$E_{nx,ny}=\frac{h^2}{8m}\Big(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\Big),$
where $m$ is the electron mass and $n_x$ is a quantum number for well width $L_x$ and $n_y$ is a quantum number for well width $L_y$.
We probe for the electron along a line that bisects $L_x$ and find three points at which the detection probability is maximum. Those points are separated by $2.00\;nm$. Therefore, $n_y=3$ and $L_y=2.00\times3\;nm=6\;nm$
Again, We probe for the electron along a line that bisects $L_y$ and find five points at which the detection probability is maximum. Those points are separated by $3.00\;nm$. Therefore, $n_x=5$ and $L_x=3.00\times5\;nm=15\;nm$
Therefore, The energy of the electron in the given two dimensional infinite well is
$E_{5,3}=\frac{(6.63\times 10^{-34})^2}{8\times 9.1\times 10^{-31}}\Big\{\frac{5^2}{(15\times 10^{-9})^2}+\frac{3^2}{(6\times 10^{-9})^2}\Big\}$
$E_{5,3}\approx 2.18\times 10^{-20}\;J$
$E_{5,3}\approx 0.136\;eV$
