Initially, when the sphere is rolling down the include without any slipping, then let the angular velocity of the sphere at the bottom of the incline be w.
No slipping condition, V = wR
Furthermore, due to conservation of energy
PE lost = KE gained
mgh = 1/2mv^2 + 1/2 *I*w^2 (Where h is the height of the incline and I is the moment of inertia of sphere around its center)
Also, I = m*Rg ^2 where Rg is the radius of gyration and also v = wR (No slipping condition)
Hence, mgh = 1/2mv^2 + 1/2 *m*Rg ^2 *v^2 /R^2
2gh = [1+(Rg/R)^2]*v^2 (1)
Furthermore, under no rolling case
mgh = 1/2m*(5v/4)^2
v^2 = 32/25 gh
Putting it in the equation 1
2gh = [1+(Rg/R)^2]*32/25 gh
Hence, Rg = 3/4R
So, the radius of gyration of the sphere around its center is 3/4R
