Answer
Please see the work below.
Work Step by Step
We know that
$\omega=2\pi(\frac{200}{60})=20.94\frac{rad}{s}$
We also know that the force of friction should be equal to the required centripetal force
$\mu mg=mr\omega^2$
This can be rearranged as
$r=\frac{\mu g}{\omega^2}$
We plug in the known values to obtain:
$r=\frac{(1.2)(9.8)}{(20.94)^2}=0.027m$
