Answer
The energy difference between second and third excited states is
$\frac{h^2}{8mL^2}$.
Work Step by Step
The quantized energies for an electron trapped in a two-dimensional infinite potential well that forms a rectangular corral 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 $L_y$.
Substituting, $L_x=L$, and $L_y=2L$, we get
$E_{nx,ny}=\frac{h^2}{8mL^2}\Big(n_x^2+\frac{n_y^2}{2^2}\Big),$
For the given two dimensional rectangular corral, $(1,3)$ and $(2,1)$ states are the electron's second and third excited states respectively. The corresponding energies are
$E_{1,3}=\frac{h^2}{8mL^2}\Big(1^2+\frac{3^2}{2^2}\Big)$
or, $E_{1,3}=3.25\times\frac{h^2}{8mL^2}$
and
$E_{2,1}=\frac{h^2}{8mL^2}\Big(2^2+\frac{1^2}{2^2}\Big)$
or, $E_{2,1}=4.25\times\frac{h^2}{8mL^2}$
$\therefore$ The energy difference between second and third excited states is
$E_{2,1}-E_{1,3}=\frac{h^2}{8mL^2}$.
