Given, $I=2 \,kg - m ^{2}, \omega_{0}=\frac{60}{60} \times 2 \pi\, rad / s$
$\omega=0, t=60\, s$
The torque required to stop the wheel's rotation is
$\tau=I \alpha=I\left(\frac{\omega_{0}-\omega}{t}\right) $
$\therefore \tau=\frac{2 \times 2 \pi \times 60}{60 \times 60}$
$=\frac{\pi}{15} N - m$