Answer
$I= 1.9\times10^{-47}kg \cdot m^2$
Work Step by Step
From figure 29–16, we see that a rotational absorption spectrum exhibits peaks at energies of $\frac{\hbar^2}{I}$, 2 times that, 3 times that, etc. In other words, adjacent peaks are separated by an energy of $\frac{\hbar^2}{I}$. Use the photon frequency corresponding to that energy, and calculate the rotational inertia.
$$\Delta E = \frac{\hbar^2}{I}$$
$$I= \frac{\hbar^2}{\Delta E}= \frac{\hbar^2}{hf}= \frac{h}{4 \pi^2f}$$
$$I= \frac{(6.63\times10^{-34}J \cdot s)}{4 \pi^2(8.9\times10^{11}Hz)}$$
$$I= 1.9\times10^{-47}kg \cdot m^2$$