Answer
6.16 MeV.
Work Step by Step
For a particle in a cubical box of side length L, with quantum numbers $(n_x, n_y, n_z)$, the energy is $\frac{(n_x^2+ n_y^2+n_z^2)\pi^2\hbar^2}{2mL^2}$
Therefore $E_{1,1,1}= \frac{3\pi^2\hbar^2}{2mL^2}$.
We also see that $E_{2,1,1}= \frac{6\pi^2\hbar^2}{2mL^2}$.
$\Delta E= \frac{3\pi^2\hbar^2}{2mL^2}$
Evaluating this expression, we find $\Delta E = 6.16 MeV$.