Mass of the ring $M=\rho L$. Let $R$ be the radius of the ring, then

$$ L=2 \pi R \quad \text { or } \quad R=\frac{L}{2 \pi} $$

The moment of inertia about an axis passing through the centre $O$ and perpendicular to plane of the loop is $I_z=m R^2$. Using symmetry and perpendicular axis theorem, $I_x+I_y=I_z$, we get

$$ I_x=I_y=\frac{1}{2} I_z=\frac{1}{2} m R^2, $$

where $I_x$ is the moment of inertia about an axis passing through $O$ and lying in the plane of the loop. The centre of mass $O$ of the loop lies at a perpendicular distance $d=R$ from the axis $\mathrm{XX}^{\prime}$. The parallel axis theorem gives

$$ I_{X X^{\prime}}=\frac{1}{2} M R^2+M R^2=\frac{3}{2} M R^2 $$

Substituting values of $M$ and $R$

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