Answer
The minimum speed at the top of the loop is 2.42 m/s.
Work Step by Step
At the minimum speed, the normal force is zero at the top of the loop and the centripetal force is provided by gravity.
$mg = \frac{mv^2}{r}$
$v = \sqrt{gr} = \sqrt{(9.80~m/s^2)(0.600~m)}$
$v = 2.42~m/s$
The minimum speed at the top of the loop is 2.42 m/s.