= Solution
For <optically thin CO rotational thermometry> of the same emitting population, the upper-level abundance ratio is
$$
\frac{N_3}{N_1}=\frac73\exp\left[-\frac{E_3-E_1}{k_BT}\right].
$$
The difference $E_3-E_1$ can be obtained directly from the two successive transition frequencies, $h(\nu_{32}+\nu_{21})$, or approximately from $10k_BT_r$. Using the supplied rounded molecular data gives $(E_3-E_1)/k_B=27.66$ K. Therefore
$$
\frac{L_{32}}{L_{10}}=\frac{7A_{32}\nu_{32}}{3A_{10}\nu_{10}}e^{-27.66/T}
=242.99\,e^{-27.66/T}.
$$
The observed ratio is $2.46\times10^{26}/(1.33\times10^{24})=184.96$, hence
$$
\boxed{T=\frac{27.66}{\ln(242.99/184.96)}\simeq101\ \mathrm K}.
$$
At 20 K the predicted ratio would be only about 61, so the two detections reject that single-temperature optically thin 20-K model. The inferred value is an excitation <temperature> under the stated assumptions, not a model-independent measurement of the gas's <kinetic temperature>.
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