A thin wire of length ' L ' and linear mass density ' m ' is bent...

Total mass of the ring, $M = m L$.

Since the wire of length $L$ is bent into a circular ring of radius $R$, its circumference is $L$.

$2\pi R = L \Rightarrow R = \dfrac{L}{2\pi}$

The axis $yy'$ is a tangent to the ring in its plane. The moment of inertia of a ring about its diametric axis is $I_{d} = \dfrac{1}{2}MR^2$.

Using the parallel axis theorem, the moment of inertia of the ring about the tangent $yy'$ is:

$I_{yy'} = I_{d} + MR^2$

$I_{yy'} = \dfrac{1}{2}MR^2 + MR^2 = \dfrac{3}{2}MR^2$

Substituting the values of $M$ and $R$:

$I_{yy'} = \dfrac{3}{2}(mL)\left(\dfrac{L}{2\pi}\right)^2$

$I_{yy'} = \dfrac{3}{2}mL \left(\dfrac{L^2}{4\pi^2}\right)$

$I_{yy'} = \dfrac{3mL^3}{8\pi^2}$

Answer: $\dfrac{3\ mL^{3}}{8\pi^{2}}$