Step 1: Recall how rotational lines are spaced.
In the rigid rotor model, the pure rotational absorption lines of a diatomic molecule sit at $\tilde{\nu}_J = 2B_e(J+1)$ for $J=0,1,2,...$, so any two neighbouring lines are separated by exactly $2B_e$, a constant independent of $J$. This is why the spectrum shows evenly spaced lines.
Step 2: Write the spacing directly in terms of I.
Since $B_e = h/(8\pi^2 c I)$, the spacing itself can be written directly in terms of the moment of inertia, without solving for $B_e$ as a separate number:
\[ \Delta\tilde{\nu} = 2B_e = \frac{h}{4\pi^2 c I} \]
Step 3: Solve for I.
Rearranging:
\[ I = \frac{h}{4\pi^2 c\, \Delta\tilde{\nu}} \]
This single expression uses the measured line spacing directly, so there is no need to compute $B_e$ as an intermediate step.
Step 4: Substitute the numbers.
With $h = 6.6\times 10^{-34}$ J.s, $c = 3\times 10^{10}$ cm/s (converted to cm/s since $\Delta\tilde\nu$ is in $\text{cm}^{-1}$), and $\Delta\tilde\nu = 20.0\ \text{cm}^{-1}$:
\[ I = \frac{6.6\times 10^{-34}}{4\pi^2 \times 3\times 10^{10}\times 20.0} = \frac{6.6\times 10^{-34}}{2.369\times 10^{13}} = 2.79\times 10^{-47}\text{ kg.m}^2 \]
Final Answer:
Rounded to one decimal place, $I_0 = 2.8$.
\[ \boxed{I_0 = 2.8} \]