A wheel having moment of inertia 2 kg–m$^2$ about its vertical axis, rotates at the rate of 60 rpm about the axis. The torque which can stop the wheel's rotation in one minute would be :-

Use $\tau = I\alpha$ and $\omega = \omega_0 + \alpha t$.
Here $\omega_0 = \dfrac{60 \times 2\pi}{60}$ rad/s $= 2\pi$ rad/s
Final $\omega = 0$ in $t = 60$ s, so the magnitude of angular retardation $\alpha = \dfrac{\omega_0}{t} = \dfrac{2\pi}{60}$ rad/s$^2$
$\tau = I\alpha = I\dfrac{\omega_0}{t} = 2 \times \dfrac{2\pi}{60}$
$\tau = \dfrac{\pi}{15}$ N–m