#### Answer

The correct answer is:
A. Less than 1

#### Work Step by Step

The retarding force of 20 N would have to do 30 J of negative work to bring the wheel to a rest. This force would need to be applied over a tangential distance of 1.5 meters, (since $Work = F~d$)
We can find the number of rotations:
$\frac{1.5~m}{2\pi~r} = \frac{1.5~m}{(2\pi)(0.3~m)} = 0.8$
The correct answer is:
A. Less than 1