The correct answer is $\frac{\pi }{{64}}\left( {{{\text{D}}^4} + {{\text{d}}^4}} \right)$.
The moment of inertia of a thin ring about an axis perpendicular to the plane of the ring is given by:
$$I = \frac{\pi }{{64}}\left( {{{\text{D}}^4} + {{\text{d}}^4}} \right)$$
where $D$ is the external diameter of the ring and $d$ is the internal diameter of the ring.
The moment of inertia is a measure of how difficult it is to change the rotation of an object. The larger the moment of inertia, the more difficult it is to rotate the object.
The moment of inertia of a thin ring is calculated by considering the mass of each infinitesimal element of the ring and its distance from the axis of rotation. The mass of each element is given by:
$$dm = \pi r \, dr$$
where $r$ is the distance from the element to the axis of rotation.
The moment of inertia of each element is given by:
$$I_d = mr^2$$
The total moment of inertia is then given by:
$$I = \int_0^D I_d \, dr = \int_0^D \pi r^3 \, dr = \frac{\pi }{{64}}\left( {{{\text{D}}^4} + {{\text{d}}^4}} \right)$$