A thin wire of length ' $L$ ' and uniform linear mass density ' $\lambda$ ' is bent into a circular ring. The moment of inertia of ring about a tangential axis in its plane is

Given a wire of length $L$ with uniform linear mass density $\lambda$, the total mass is $M = \lambda L$.

When formed into a circular ring, the circumference equals the wire length, so $L = 2\pi R$ and $R = \frac{L}{2\pi}$.

The moment of inertia about a diameter is $I_d = \frac{MR^2}{2}$ by the perpendicular axis theorem.

Applying the parallel axis theorem for a tangential axis in the plane, located distance $R$ from the center:

$I_{\text{tangential}} = I_d + MR^2 = \frac{MR^2}{2} + MR^2 = \frac{3}{2}MR^2$.

Substituting $M = \lambda L$ and $R = \frac{L}{2\pi}$:

$I_{\text{tangential}} = \frac{3}{2}(\lambda L)\left(\frac{L}{2\pi}\right)^2 = \frac{3\lambda L^3}{8\pi^2}$.

Final answer: $\boxed{\text{C}}$