Answer
See explanation.
Work Step by Step
For a particle of mass m 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 $E_{2,2,1}= \frac{9\pi^2\hbar^2}{2mL^2}$.
We also see that $E_{1,1,1}= \frac{3\pi^2\hbar^2}{2mL^2}$.
$\Delta E= \frac{6\pi^2\hbar^2}{2mL^2}$
Evaluating this expression, we find $\Delta E = 353 eV$.
The energy of the photon is equal to the energy difference between the states.
$$E= hf=h\frac{c}{\lambda} $$
$$\lambda =\frac{hc}{E}=\frac{1.24\times10^{-6}eV\cdot m}{353eV}=3.51\times10^{-9}m$$
