Answer
$I = \frac{1}{2}mR^2$
Work Step by Step
By conservation of energy, the kinetic energy at the bottom of the hill is equal to the gravitational potential energy at the top of the hill.
We can find the rotational inertia:
$K = U$
$K = mgh$
$K = mg(\frac{3v^2}{4g})$
$K = \frac{3}{4}~mv^2$
$K = \frac{1}{2}~mv^2+\frac{1}{4}~mv^2$
$K = \frac{1}{2}~mv^2+\frac{1}{4}~m(\omega~R)^2$
$K = \frac{1}{2}~mv^2+\frac{1}{2}\cdot \frac{1}{2}mR^2~\omega^2$
$K = \frac{1}{2}~mv^2+\frac{1}{2} I~\omega^2$
Therefore:
$I = \frac{1}{2}mR^2$
