Answer
$69\Big(\frac{h^2}{8mL^2}\Big)$
Work Step by Step
To solve the current problem, we have to first determine the ground state configuration of the system.
We have put $11$ electrons in the given fictitious infinite potential well . The trapped electrons must obey the Pauli exclusion principle; that is, no two electrons can have the same set of values for their quantum numbers. Thus, according to the Pauli exclusion principle, the state having lowest energy $4\Big(\frac{h^2}{8mL^2}\Big)$ will be first fully occupied with $2$ electrons. Then the triple degenerate state having energy $6\Big(\frac{h^2}{8mL^2}\Big)$ will be fully occupied with $6$ electrons. Next, the remaining $3$ electrons will go into the next double degenerate state having energy $7\Big(\frac{h^2}{8mL^2}\Big)$.
Thus the ground state energy of the system is
$E_{gr}=2\times4\Big(\frac{h^2}{8mL^2}\Big)+6\times6\Big(\frac{h^2}{8mL^2}\Big)+3\times7\Big(\frac{h^2}{8mL^2}\Big)$
or, $E_{gr}=65\Big(\frac{h^2}{8mL^2}\Big)$
If the system is to be excited, one of the $11$ 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 least energy $(\Delta E)$ is required for an electron to jump from $7\Big(\frac{h^2}{8mL^2}\Big)$ energy state to $11\Big(\frac{h^2}{8mL^2}\Big)$.
Therefore, for the system to jump from its ground state to its first excited state, the electron in $7\Big(\frac{h^2}{8mL^2}\Big)$ energy level must jump to the partially occupied $11\Big(\frac{h^2}{8mL^2}\Big)$ energy level and the required energy is
$\Delta E=11\Big(\frac{h^2}{8mL^2}\Big)-7\Big(\frac{h^2}{8mL^2}\Big)=4\Big(\frac{h^2}{8mL^2}\Big)$
The energy $E_{fe}$ of the first excited state of the system is then
$E_{fe}=E_{gr}+\Delta E$
or, $E_{fe}=65\Big(\frac{h^2}{8mL^2}\Big)+4\Big(\frac{h^2}{8mL^2}\Big)$
or, $E_{fe}=69\Big(\frac{h^2}{8mL^2}\Big)$
