= Solution
Count the successive <Q branch> lines from the low-frequency edge, taking $J=1$ as the first allowed bend transition; the first lines are crowded and partly unresolved. The maximum near $712.55\ \mathrm{cm}^{-1}$ is approximately \b[$J=12$]. Treating $J$ as continuous in the intensity envelope from part (i),
$$
\frac{\mathrm d}{\mathrm dJ}\log I_Q(J)
=\frac2{2J+1}-\frac{hcB''}{k_{\rm B}T}(2J+1)=0.
$$
Hence
$$
\boxed{T\simeq\frac{hcB''}{2k_{\rm B}}(2J_{\max}+1)^2.}
$$
Using $hc/k_{\rm B}=1.43878\ \mathrm{K\,cm}$, $B''=1.47786\ \mathrm{cm}^{-1}$ and $J_{\max}\simeq12$ gives
$$
T\simeq\frac{(1.43878)(1.47786)}2(25)^2
\simeq664\ \mathrm K.
$$
A useful check avoids locating the exact maximum. The plotted $J\simeq20$ line has roughly $0.70$ of the $J\simeq12$ peak height. Thus
$$
\frac{I_{20}}{I_{12}}\simeq\frac{41}{25}
\exp\!\left[-\frac{hcB''(420-156)}{k_{\rm B}T}\right]
$$
gives about $659\ \mathrm K$. \b[An appropriate graphical estimate is $T\simeq6.6\times10^2\ \mathrm K$], with precision limited by line counting, line overlap and reading the plotted intensities. There is no basis for reporting a temperature to the nearest kelvin from this figure.
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