= Solution
Set the ground electronic energy to zero and write the excitation energy as $\Delta=hc(120\ \mathrm{cm^{-1}})$, or the characteristic temperature $\theta=\Delta/k_{\mathrm B}=172.65$ K. Counting each doubly degenerate manifold once gives the single-molecule <electronic partition function>
$$
\boxed{q_{\mathrm{el}}=2+2e^{-\beta\Delta}=2(1+e^{-\theta/T}).}
$$
For $N$ independent molecules the electronic factor is $q_{\mathrm{el}}^N$. Hence the electronic <internal energy> is
$$
\boxed{U_{\mathrm{el}}=-N\frac{\partial\ln q_{\mathrm{el}}}{\partial\beta}
=\frac{N\Delta e^{-\beta\Delta}}{1+e^{-\beta\Delta}}
=\frac{N\Delta}{1+e^{\theta/T}}.}
$$
The equal degeneracies cancel from the excited-state probability. At low temperature the ground manifold dominates and $U_{\mathrm{el}}\to0$; at high temperature both manifolds have equal populations, giving $U_{\mathrm{el}}\to N\Delta/2$. These are electronic contributions only, excluding translation, rotation and vibration.
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