= Solution
For one mole, $Nk=R$, so the three-level <Morse oscillator> result is
$$
\boxed{U_m=RT^2\frac{q_3'}{q_3}=124.21\ \mathrm{J\,mol^{-1}}}.
$$
For the full <quantum harmonic oscillator> with the given wavenumber,
$$
\boxed{U_{m,\mathrm{harm}}=\frac{R(215/0.695)}{e^{215/(0.695\cdot100)}-1}
=122.16\ \mathrm{J\,mol^{-1}}}.
$$
The specified anharmonic calculation is about $1.7\%$ higher. Anharmonicity lowers the first excitation spacings and increases excited-state occupation. Both results remain small because $100\ \mathrm K$ is well below the vibrational <temperature>. The comparison also mixes a three-level anharmonic truncation with an infinite harmonic sum; omitted higher Morse levels would make a further small positive correction at this <temperature>.
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