= Solution
Using $Z_{\rm rot}\simeq CT=C/(k_B\beta)$ gives
$$
\ln Z_{\rm rot}=\ln\left(\frac{C}{k_B\beta}\right).
$$
The internal-energy identity from (b), followed by <differentiation> with respect to <temperature> at fixed <moment of inertia>, yields
$$
\boxed{U_{\rm rot}\simeq\frac1\beta=k_BT,\qquad C_{V,\rm rot}=\left(\frac{\partial U_{\rm rot}}{\partial T}\right)_I\simeq k_B.}
$$
For one mole of independent rotors, the rotational <energy> is $RT$ and the rotational molar <heat capacity> is $R\simeq8.31\,\mathrm{J\,mol^{-1}\,K^{-1}}$.
This agrees with the <equipartition theorem>. The given spectrum describes a linear rotor with two independent rotational <degrees of freedom>, each contributing $k_BT/2$ to the high-temperature mean <energy>. Rotation about the molecular axis is not an additional <degree of freedom> of this ideal linear-rotor model, so the result is $k_B$ rather than $3k_B/2$ per rotor. These are rotational contributions alone; translational and vibrational <heat capacities> have not been included. At low <temperature> the discrete excitation gap freezes rotational excitation out, so the rotational <heat capacity> tends to zero instead of remaining at its classical value.
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