Answer
$\beta = 0.50$
Work Step by Step
Note that the magnitude of the normal force $F_N$ at the bottom of the ramp is equal to the sum of the centripetal force and the gravitational force.
Let $r$ be the radius of the circular loop.
We can find an expression for $v^2$ of the ball at the bottom of the ramp:
$F_N = \frac{Mv^2}{r} +Mg = 2.00~Mg$
$\frac{v^2}{r} = g$
$v^2 = g~r$
We can find an expression for the total kinetic energy at the bottom of the ramp:
$K_2 = \frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$
$K_2 = \frac{1}{2}Mv^2+\frac{1}{2}(\beta~MR^2)(\frac{v}{R})^2$
$K_2 = \frac{1}{2}Mv^2+\frac{\beta}{2}Mv^2$
$K_2 = \frac{\beta+1}{2}Mv^2$
$K_2 = \frac{\beta+1}{2}M~g~r$
Note that the total kinetic energy at the bottom of the ramp is equal to the gravitational potential energy at $h = 0.36~m$
We can use conservation of energy to find $\beta$:
$K_2+U_2 = K_1+U_1$
$K_2+0 = 0+U_1$
$\frac{\beta+1}{2}M~g~r = Mgh$
$\frac{\beta+1}{2}~r = h$
$\beta+1 = \frac{2h}{r}$
$\beta = \frac{2h}{r}-1$
$\beta = \frac{(2)(0.36~m)}{0.48~m}-1$
$\beta = 0.50$
