Answer
$\omega=14.85rad/s$
Work Step by Step
The stone is released horizontally, so its initial vertical speed is 0.
As the stone falls vertically, we have $v_0=0, g=9.8m/s^2$ and $s=20m$. The time it takes the stone to reach the ground is $$s=v_0t+\frac{1}{2}gt^2=0+\frac{1}{2}gt^2$$ $$t=\sqrt{\frac{2s}{g}}=2.02s$$
We call the speed the stone is released horizontally $v$. During $\Delta t=2.02s$, it travels a horizontal distance of $$X=2.02v (1)$$
2) The stone's rotational motion $v$ is the stone's tangential velocity at the time of release. We call the whirling circle's radius $r$ and have $$\omega=\frac{v}{r}$$
From (1), we get $v=X/2.02$ $$\omega=\frac{X}{2.02r}$$
The exercise also claims $X=30r$, so $$\omega=\frac{30r}{2.02r}=14.85rad/s$$
