Interpret the given quantity carefully

The ring has:

• mass $M$
• diameter $r$

Hence its radius is $R=\frac{r}{2}.$

Moment of inertia of a ring about a diameter in its plane

For a thin circular ring, the moment of inertia about the axis perpendicular to its plane through the center is $I_z=MR^2.$

By the perpendicular axis theorem, $I_x+I_y=I_z.$

Since the ring is symmetric, $I_x=I_y.$

Therefore, about any diameter in the plane, $I_{\text{diameter}}=\frac{1}{2}MR^2.$

Use parallel axis theorem for tangential axis in the plane

The required axis is tangential to the ring and lies in its plane. This axis is parallel to a diameter and at a distance $R$ from the center.

So by parallel axis theorem, $I_{\text{tangent}}=I_{\text{diameter}}+MR^2.$

Substituting, $I_{\text{tangent}}=\frac{1}{2}MR^2+MR^2=\frac{3}{2}MR^2.$

Express in terms of the given diameter $r$

Since $R=\frac{r}{2}$,

Match with the options

$\boxed{I=\frac{3}{8}Mr^2}$

So the correct option is A.

Comparison with stored answer

Stored correct answer: A

Our derived answer: A

Hence, they agree.