Right option is (b) 25/3π N
The best I can explain: Let w be the angular velocity & a be the angular acceleration.
Moment of inertia of the ring = MR^2
= 2*0.25 = 0.5kgm^2.
Angular distance = 3 rotations = 3*2π rad = 6π rad.
Using, w^2 = w0^2 + 2aθ, we get:
0 = 100 – 2a*6π.
a = 100/12π.
Torque = Ia = rF
∴ F = Ia/r = 0.5*100/(12π*0.5) = 25/3π N.