A thin wire of length L and uniform linear mass density $\rho$ is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis XX' is

Mass of the ring  $\mathrm{M}=\mathrm{\rho L}$
Let R be the radius of the ring, then
$L=2\mathrm{\pi R}\Rightarrow R=\frac{L}{2\mathrm{\pi }}$
Moment of inertia about an axis passing through O and parallel to  $X{X}^{\text{'}}$will be  $I_0=\frac{1}{2}M{R}^2$

Therefore, moment of inertia about  $X{X}^{\text{'}}$ (from parallel axis theorem) will be given by  $I_{X{X}^{\text{'}}}=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

Substituting values of M and R:

$I_{X{X}^{\text{'}}}=\frac{3}{2}\left(\rho L\right)\left(\frac{L^2}{4{\pi }^2}\right)=\frac{3\rho L^3}{8{\pi }^2}$