Consider a diatomic molecule along $z$-axis is so its rotational energy about $z$-axis is zero.
So, the total energy associated with the diatomic molecule is
$$
E=\frac{1}{2} m v_x^2+\frac{1}{2} m v_y^2+\frac{1}{2} m v_z^2+\frac{1}{2} I_x \omega_x^2+\frac{1}{2} I_y \omega_y^2
$$
In the above expression the independent term is $(5)$ and it contains translational kinetic energy $\left(\frac{1}{2} m v^2\right)$ corresponding to velocity in each $x, y$ and $z$-directions as well as rotation $\mathrm{KE}\left(\frac{1}{2} I \omega^2\right)$ associated with axis of rotations $x$ and $y$.


As we can predict velocities of molecules by Maxwell's distribution, hence the above expression also obeys Maxwell's distribution.
As 2 rotational and 3 translational energies are associated with each molecule.
So the rotational energy at a given temperature is $\left(\frac{2}{3}\right)$ rd of its translational KE of each molecule.