Answer
$48\Big(\frac{h^2}{8mL^2}\Big)$
Work Step by Step
From the problem $23$, the ground state energy $E_{gr}$ is the sum of the energies of the individual electrons in the system’s ground-state configuration. Therefore,
$E_{gr}=2E_{1,1,1}+2E_{2,1,1}+2E_{1,2,1}+2E_{1,1,2}$
or, $E_{gr}=2\times3\Big(\frac{h^2}{8mL^2}\Big)+2\times6\Big(\frac{h^2}{8mL^2}\Big)+2\times6\Big(\frac{h^2}{8mL^2}\Big)+2\times6\Big(\frac{h^2}{8mL^2}\Big)$
or, $E_{gr}=42\Big(\frac{h^2}{8mL^2}\Big)$
If the system is to be excited, one of the $8$ electrons must make a quantum jump from one state to another. The Pauli exclusion principle must still apply. Therefore, the electrons must jump to either empty or partially occupied states.
Among the possible jumps, the third least energy $(\Delta E)$ is required for an electron to jump from the lowest $E_{1,1,1}$ energy level to 3-fold degenerate $E_{2,2,1}$ energy level.
Therefore, for the system to jump from its ground state to its third excited state, the electron in $E_{1,1,1}$ energy level must jump to unoccupied $E_{2,2,1}$ energy level and the required energy is
$\Delta E=E_{2,2,1}-E_{1,1,1}$
or, $\Delta E=9\Big(\frac{h^2}{8mL^2}\Big)-3\Big(\frac{h^2}{8mL^2}\Big)=6\Big(\frac{h^2}{8mL^2}\Big)$
The energy $E_{te}$ of the third excited state of the system is then
$E_{te}=E_{gr}+\Delta E$
or, $E_{te}=42\Big(\frac{h^2}{8mL^2}\Big)+6\Big(\frac{h^2}{8mL^2}\Big)$
or, $E_{te}=48\Big(\frac{h^2}{8mL^2}\Big)$
