= Solution
Subtracting the $v=0$ energy in the <Morse oscillator> gives
$$
\frac{\varepsilon_v-\varepsilon_0}{hc}
=v\widetilde\omega-v(v+1)\widetilde\omega x.
$$
The first three relative wavenumbers are $0$, $213$, and $424\ \mathrm{cm}^{-1}$. Thus, with $\theta_1=213/0.695\ \mathrm K$ and $\theta_2=424/0.695\ \mathrm K$,
$$
\boxed{q_3(T)=1+e^{-\theta_1/T}+e^{-\theta_2/T}},\qquad
\boxed{q_3(100\ \mathrm K)=1.04891}.
$$
This is a three-level <partition function>, as stipulated, rather than an unrestricted sum of the anharmonic formula to arbitrarily large $v$.
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