Answer
$2\Big(\frac{h^2}{8mL^2}\Big)$
Work Step by Step
The quantized energies for an electron trapped in a three-dimensional infinite potential well that forms a e rectangular box are
$E_{nx,ny,nz}=\frac{h^2}{8m}\Big(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2}\Big)$,
where m is the electron mass and $n_x$ is a quantum number for well width$ L_x$, $n_y$ is a quantum number for well width $L_y$ and $n_z$ is a quantum number for well width $L_z$.
In this problem, the electron is contained in a cubical box of widths $L_x=L_y=L_z=L$.
$\therefore\;\;E_{nx,ny,nz}=\frac{h^2}{8mL^2}\Big(n_x^2+n_y^2+n_z^2\Big)$,
The second exited states are the $(2,2,1)$, $(1,2,2)$ and $(2,1,2)$ states having same energy. Therefore, the energy level corresponds to the second excited states is $3$-fold degenerate and the energy is
$E_{2,2,1}=E_{1,2,2}=E_{2,1,2}=\frac{h^2}{8mL^2}\Big(2^2+2^2+1^2\Big)$
$\implies E_{2,2,1}=E_{1,2,2}=E_{2,1,2}=9\Big(\frac{h^2}{8mL^2}\Big)$
The third exited states are the $(3,1,1)$, $(1,3,1)$ and $(1,1,3)$ states having same energy. Therefore, the energy level corresponds to the third excited state is $3$-fold degenerate and the energy is
$E_{3,1,1}=E_{1,3,1}=E_{1,1,3}=\frac{h^2}{8mL^2}\Big(3^2+1^2+1^2\Big)$
$\implies E_{3,1,1}=E_{1,3,1}=E_{1,1,3}=11\Big(\frac{h^2}{8mL^2}\Big)$
Therefore, the difference between the energies of its second and third excited states is
$\Delta E=E_{3,1,1}-E_{2,2,1}=2\Big(\frac{h^2}{8mL^2}\Big)$
