When a ring moves by rotation without slipping on a surface the highest points P of the ring will have a speed `2v` and the lowest points will have the speed 0. Thus the highest points of the ring will rotate about the lowest points of the ring with an angular frequency `omega` and a radius of gyration `2R` .

`2v =omega*2R`

It is like when the motion of the upper points is composed of a linear translation with speed `v` of the center of ring and a rotation about the center of the ring with the same angular speed `omega` (see the figure below)

`v_("tot") = v +omega*R = v+v =2v`

Therefore the radius of the curvature of a particle at the highest point of the ring is `2R` . **The correct answer is B**.