To find the quantized rotational energy of a diatomic molecule, we can use Bohr's quantization condition for angular momentum. Bohr's quantization condition states that the angular momentum $ L $ of an electron in a hydrogen atom is quantized and given by:

$$ L = n \hbar $$

where $ n $ is a positive integer (known as the quantum number) and $ \hbar $ is the reduced Planck's constant, given by $ \hbar = {\hbar \over {2\pi}} $.

For a diatomic molecule, we extend this concept to its rotational motion. A rigid diatomic molecule can be approximated as a rigid rotor with a moment of inertia $ I $. The angular momentum $ L $ for the molecule in the nth level is given by:

$$ L = n \hbar $$

where $ n $ is the quantum number (note that $ n = 0 $ is not allowed).

The rotational kinetic energy for a rotating body with moment of inertia $ I $ is given by:

$$ E = {1 \over 2} I \omega^2 $$

where $ \omega $ is the angular velocity. The angular momentum $ L $ is related to the angular velocity $ \omega $ by:

$$ L = I \omega $$

Solving for $ \omega $, we get:

$$ \omega = {L \over I} $$

Substituting this back into the expression for energy, we get:

$$ E = {1 \over 2} I \left( {L \over I} \right)^2 = {L^2 \over 2I} $$

Using Bohr's quantization condition $ L = n \hbar $, the energy expression becomes:

$$ E = {{(n \hbar)^2} \over {2I}} $$

Substituting $ \hbar = {\hbar \over {2\pi}} $, we obtain:

$$ E = {n^2 \over {2I}} \left( {{\hbar^2} \over {4\pi^2}} \right) = n^2 \left( {{\hbar^2} \over {8\pi^2 I}} \right) $$

Therefore, the quantized rotational energy of the diatomic molecule in the nth level is:

$$ E_n = n^2 \left( {{\hbar^2} \over {8\pi^2 I}} \right) $$

So, the correct option is:

Option D

$$ n^2 \left( {{{h^2} \over {8\pi^2 I}}} \right) $$