Answer
$ T=2\pi\sqrt{\frac{2R}{g}}$
Work Step by Step
The time period of the hoop can be determined as
$T=2\pi\sqrt{\frac{l}{g}}\sqrt{\frac{I}{ml^2}}$
Substituting $l=R$ and $I=2mR^2$, we obtain:
$T=2\pi\sqrt{\frac{R}{g}}\sqrt{\frac{2mR^2}{mR^2}}$
$\implies T=2\pi\sqrt{\frac{2R}{g}}$
