Answer
$186\Big(\frac{h^2}{8mL^2}\Big)$
Work Step by Step
We have put $22$ electrons in the given fictitious three-dimensional
infinite potential. 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.
In the ground state, the 22 electrons will be arranged in the following ways according to the Pauli exclusion principle
First, the state having lowest energy $3\Big(\frac{h^2}{8mL^2}\Big)$ 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.
Then triple degenerate state having energy $9\Big(\frac{h^2}{8mL^2}\Big)$ will again be fully occupied with $6$ electrons.
Then, the triple degenerate state having energy $11\Big(\frac{h^2}{8mL^2}\Big)$ will also be fully occupied with $6$ electrons.
Then, the remaining $2$ electrons will go into the next non-degenerate state having energy $12\Big(\frac{h^2}{8mL^2}\Big)$.
Thus the ground state energy of the system is
$E_{gr}=2\times3\Big(\frac{h^2}{8mL^2}\Big)+6\times6\Big(\frac{h^2}{8mL^2}\Big)+6\times9\Big(\frac{h^2}{8mL^2}\Big)+6\times11\Big(\frac{h^2}{8mL^2}\Big)+2\times12\Big(\frac{h^2}{8mL^2}\Big)$
or, $E_{gr}=186\Big(\frac{h^2}{8mL^2}\Big)$
