= Solution
\b[$I_{C_4}\simeq2.43\times10^{-45}\ \mathrm{kg\,m^2}=146.6\,m_{\rm u}\mathring{\mathrm A}^{,2}$.] The parallel-band progression gives $2\widetilde B=0.42\ \mathrm{cm}^{-1}$, so $\widetilde B=0.210\ \mathrm{cm}^{-1}$. The perpendicular Q-feature spacing gives
$$
2|\widetilde B-\widetilde C|=0.19\ \mathrm{cm}^{-1}.
$$
For an <oblate symmetric top>, $I_{C_4}>I_\perp$, hence $\widetilde C<\widetilde B$ and $\widetilde C=0.115\ \mathrm{cm}^{-1}$. The <rotational constant> relation then gives
$$
I_{C_4}=\frac{h}{8\pi^2c\widetilde C}
=\frac{h}{8\pi^2(2.9979\times10^{10}\ \mathrm{cm\,s^{-1}})(0.115\ \mathrm{cm}^{-1})}.
$$
The estimated constants neglect centrifugal distortion and the change of rotational constants in the excited vibration.
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