Answer
0.0976 eV.
Work Step by Step
The energy levels are given by $E_n=(n+\frac{1}{2})\hbar\omega$.
The ground state energy is when n=0.
The energy of the photon equals the energy difference between the two states. Find the energy.
$$\Delta E=\hbar \omega=\frac{hc}{\lambda}$$
$$\Delta E =\frac{1.24\times10^{-6}eV\cdot m}{6.35\times10^{-6}m }=0.195eV$$
The ground state energy is $ \frac{1}{2}\hbar\omega$, which is half the spacing between energy levels.
$$ \frac{1}{2}\hbar\omega=0.0976 eV $$
