= Solution
In the rigid-rotor approximation, $E_J=\hbar^2J(J+1)/(2I)$ and electric-dipole absorption has $\Delta J=+1$. A line labelled $J'$ therefore has
$$
\nu_{J'}=\frac{hJ'}{4\pi^2I},\qquad
\Delta\nu=\frac h{4\pi^2I}.
$$
The original graph gives adjacent spacing about $0.115\,\mathrm{THz}$; for example the $J'=5$ line is near $0.575\,\mathrm{THz}$. Thus $I\simeq1.46\times10^{-46}\,\mathrm{kg\,m^2}$. Using the <reduced mass>,
$$
\mu=\frac{12\times16}{12+16}m_H=1.145\times10^{-26}\,\mathrm{kg},
\qquad
\boxed{r=\sqrt{I/\mu}\simeq1.13\times10^{-10}\,\mathrm m=0.113\,\mathrm{nm}.}
$$
<Centrifugal distortion> of a nonrigid molecule explains small departures from equal spacing at higher $J$.
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