Answer
See explanation.
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 $\frac{(n_x^2+ n_y^2+n_z^2)\pi^2\hbar^2}{2mL^2}$.
Here, we are told that $L^3=\frac{4\pi a_o^3}{3}$, where $a_o$ is the Bohr radius. This means that $L=8.53\times10^{-11}m$.
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 $\Delta E = 155 eV$.
In contrast, in the Bohr model, $E_n=-\frac{13.6eV}{n^2}$ so the energy difference between the first and second levels is $\Delta E =-13.6eV (\frac{1}{2^2}-\frac{1}{1^2})=10.2 eV$.
We conclude that a particle in a box is not a great model for a hydrogen atom.
