For a circular wire (thin ring) of mass $M$ and radius $R$, the moment of inertia about an axis passing through its center and perpendicular to the plane is $I_z = MR^2.$

We need the moment of inertia about a diameter of the ring, i.e. an axis lying in the plane of the ring and passing through its center.

By the perpendicular axis theorem for a plane lamina, $I_z = I_x + I_y$ where $I_x$ and $I_y$ are the moments of inertia about two perpendicular diameters in the plane.

Since the ring is symmetric, the moment of inertia about any diameter is the same. Hence, $I_x = I_y.$ Let this common value be $I_d$. Then, $MR^2 = I_d + I_d = 2I_d.$

Therefore, $I_d = \frac{MR^2}{2}.$

So the correct option is $\boxed{\frac{MR^2}{2}}$ which corresponds to Option A.