Using Bohr's quantization condition, the rotational kinetic energy in the third orbit for a diatomic molecule is ( $h=$ Planck's constant, $I=$ moment of inertia of diatomic molecule)

Rotational kinetic energy in the third orbit follows from Bohr's quantization of angular momentum: $L = n \frac{h}{2\pi}$, where $n = 3$ for the third orbit, giving $L_3 = \frac{3h}{2\pi}$.

Substituting into the rotational kinetic energy formula $KE_{\text{rot}} = \frac{L^2}{2I}$ yields:

$KE_{\text{rot}} = \frac{\left(\frac{3h}{2\pi}\right)^2}{2I} = \frac{\frac{9h^2}{4\pi^2}}{2I} = \frac{9h^2}{8\pi^2 I}$.

This corresponds to option A.