Since the wheel start from rest. angular acceleration,
θ=1/2αt²
14=1/2α x 8.7²
α= 0.3699 rad/s²
moment of inertia of loop= mr²= 4.1x0.37=1.517 kgm²
torque=T= lα
T= 0.5611Nm= 0.56Nm to significant figure
Disc
moment of inertia of disc= 1/2mr²
Required torque value= 0.28Nm
So,
I= 1/2X 4.1X 0.37²= 0.280 Kgm²
T= Iα = 0.280 X 0.3699= 0.10 to two significant figure