Answer
$\omega = \frac{mvR}{I+MR^2}$
Work Step by Step
Note that the initial angular momentum of the system is zero.
We can use conservation of angular momentum to find the angular speed of the merry-go-round:
$L_f = L_i$
$(I+MR^2)~\omega - mvR = 0$
$(I+MR^2)~\omega = mvR$
$\omega = \frac{mvR}{I+MR^2}$
