Moment of inertia of rod rotating about an axis passing through one end is $I_1=\frac{m{l}^2}{3} \cdots \left(1\right)$ 

Moment of inertia of ring rotating about one of its diameters is $I_2=\frac{m{R}^2}{2}$.

Since the rod is bent into a ring, the circumference of the ring and the length of the rod are equal, $l=2\pi R$

Then, $I_2=\frac{m{R}^2}{2}=\frac{m{l}^2}{8{\pi }^2}$ 

Thus, the ratio $\frac{I_1}{I_2}=\frac{8{\pi }^2}{3}$, so $x=8$.