= Solution
Define $\theta_{\rm rot}=\hbar^2/(2Ik_{\rm B})$. Summing the spatial states counted in part (c),
$$
q_{\rm planar}^{\rm labelled}=\sum_{m=-\infty}^{\infty}e^{-m^2\theta_{\rm rot}/T}
=1+2\sum_{J=1}^{\infty}e^{-J^2\theta_{\rm rot}/T}.
$$
When $T\gg\theta_{\rm rot}$ the closely spaced terms can be replaced by a <Gaussian integral>:
$$
\boxed{q_{\rm planar}^{\rm labelled}\simeq\int_{-\infty}^{\infty}
e^{-m^2\theta_{\rm rot}/T}\,dm
=\sqrt{\frac{\pi T}{\theta_{\rm rot}}}
=\frac{\sqrt{8\pi^3Ik_{\rm B}T}}h.}
$$
Here labelled means that the two ends of the molecular axis were distinguished in the angular count. In the usual high-temperature molecular convention that removes equivalent orientations and factors out nuclear-spin multiplicity, divide by the <rotational symmetry number> $s=2$ for homonuclear <hydrogen>:
$$
\boxed{q_{\rm rot}^{\rm planar}\simeq\frac1s\sqrt{\frac{\pi T}{\theta_{\rm rot}}},\qquad s=2.}
$$
The full exact <spin isomers of hydrogen> count is instead the even-$m$ spatial sum plus three times the odd-$m$ sum; its high-temperature limit is $2\sqrt{\pi T/\theta_{\rm rot}}$, including the four nuclear-spin states. These conventions differ only by temperature-independent factors in this limit. All give $U_{\rm rot,m}=RT/2$ and $C_{V,m}^{\rm rot}=R/2$, consistent with one quadratic rotational degree of freedom. By comparison, the free <linear rigid rotor> has $q_{\rm rot}\propto T$ and $C_{V,m}^{\rm rot}=R$.
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