= Solution
The relevant allowed absorption is $1\to2$. Under the stipulated population-only intensity model, the vibrational levels are nondegenerate and
$$
\frac{I_{12}}{I_{01}}=\frac{N_1}{N_0}
=\exp\left[-\frac{E_1-E_0}{k_{\mathrm B}T}\right]
=\exp\left[-\frac{hc(440.6\ \mathrm{cm^{-1}})}{k_{\mathrm B}T}\right].
$$
Set this ratio to $0.10$ and solve:
$$
\boxed{T=\frac{hc(440.6\ \mathrm{cm^{-1}})}{k_{\mathrm B}\ln10}
=\frac{(1.4387769\ \mathrm{K\,cm})(440.6\ \mathrm{cm^{-1}})}{\ln10}
\simeq275.3\ \mathrm K.}
$$
The energy in the exponential is the measured $0\to1$ gap, not the harmonic constant $444.3\ \mathrm{cm^{-1}}$. At this temperature, $N_2/N_0\simeq0.0102$, so considering the first thermally excited manifold is sensible. Actual band intensity ratios also involve transition moments and rotational structure; the calculation follows the question's simplifying assumption.
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