Correct Answer - B
Moment of inertia at the centre and perpendicular to the plane of the ring.
So, about la point on the rim of the rind and the axisw perpendicular to the plane of the rign the moment of inertia
`=mR^2+mR^2=2mR^2`
(parallel axis therom)
`rarr mK^2=2mR^2`
`(K=radis of the gyration)`
`rarr K=sqrt(2R^2)=sqrt2R`