C

To find the equivalent length of a simple pendulum for a ring of radius R suspended from its circumference, we first consider the moment of inertia and the forces acting on the ring.
Step 1: For a ring, the moment of inertia about the diametrical axis is given by \( I = \frac{1}{2} m R^2 \).
Step 2: The torque due to gravity when the ring is displaced by an angle \( \theta \) can be analyzed. The center of mass will move downward effectively creating a torque.
Step 3: The angular frequency \( \omega \) for small oscillations is given by: \( \omega = \sqrt{\frac{g}{L}} \), where L is the effective length.
Step 4: By equating the expressions for \( \omega \) derived from the motion and adjusting for the ring's geometry leads to an effective length of oscillation as \( L = \frac{3R}{2} \).
Therefore, the correct answer is \( \frac{3R}{2} \). Hence, option C is correct.