If ${\mathrm{I}}_1$ be the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and ${\mathrm{I}}_2$ be the moment of inertia of the ring formed by bending the rod, then the ratio ${\mathrm{I}}_1$:${\mathrm{I}}_2$ is

Let L represent the rod's length.

${\mathrm{l}}_1=\frac{{\mathrm{mL}}^2}{12}$

Rod bending will result in a ring with a mass of m and a radius of

$2\mathrm{\pi r}=\mathrm{L}$

$\mathrm{r} = \frac{\mathrm{L}}{2\mathrm{\pi }}$

An axis perpendicular to a ring's plane and traveling through its centre has the following moment of inertia:

${\mathrm{l}}_2={\mathrm{mr}}^2=\frac{{\mathrm{mL}}^2}{4{\mathrm{\pi }}^2}$

$\frac{{\mathrm{l}}_1}{{\mathrm{l}}_2}=\frac{{\mathrm{\pi }}^2}{3}$

Hence the correct answer is ${\mathrm{\pi }}^2:3.$