Finding the rotational inertia of a rolling ball on a ramp

A body of radius 0.323 m and mass 1.64 kg is rolling smoothly with speed 3.47 m/s on a horizontal surface. It then rolls up a hill to a maximum height 2.64 m. What is the body's rotational inertia about the rotational axis through its center of mass?

mgh = 1.64 x 9.8 x 2.64 = 42.43

linear ke = 0.5 x 1.64 x 3.47^2 = 9.874 J

42.43 - 9.874 = rotational energy = 32.556 J

RKE = 1/2Iω^2

ω = v/r = 3.47/0.323 = 10.743 rad/s

10.743^2 = 115.413

32.556/115.413 = 1/2I = 0.282

0.282 x 2 = I = 0.564 kg.m^2

for rotational inertia it only depends on the mass and radius of the object. so inertia is given by
I = xmr^2 where x is some fraction that depends on the shape of the object. sorry but i dont remember the fraction for that shape you describe.