Answer
$\beta = 0.4$
Work Step by Step
We can find the time it takes the object to fall $h = 0.10~m$:
$h = \frac{1}{2}at^2$
$t = \sqrt{\frac{2h}{a}}$
$t = \sqrt{\frac{(2)(0.10~m)}{9.8~m/s^2}}$
$t = 0.14286~s$
We can find the required horizontal speed $v$ in order for the object to travel a horizontal distance of $0.506~m$ in this time:
$x = v~t$
$v = \frac{x}{t}$
$v = \frac{0.506~m}{0.14286~s}$
$v = 3.542~m/s$
The rotational inertia of the object is $I = \beta MR^2$
We can find a general expression for the total kinetic energy:
$K = \frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$
$K = \frac{1}{2}Mv^2+\frac{1}{2}(\beta MR^2)(\frac{v}{R})^2$
$K = \frac{1}{2}Mv^2+\frac{\beta}{2}Mv^2$
$K = \frac{\beta +1}{2}Mv^2$
The gravitational potential energy at height $H$ is equal to the total kinetic energy just before rolling off the ramp.
We can use conservation of energy to find $\beta$:
$K_2+U_2 = K_1+U_1$
$K_2+0 = 0+U_1$
$\frac{\beta +1}{2}Mv^2 = MgH$
$\beta +1 = \frac{2gH}{v^2}$
$\beta = \frac{2gH}{v^2}-1$
$\beta = \frac{(2)(9.8~m/s^2)(0.90~m)}{(3.542~m/s)^2}-1$
$\beta = 0.4$
