A body rotating at 20 rad/s is subjected to a constant torque that...

The initial kinetic energy of the rotating body is given by $K_i = \dfrac{1}{2} I \omega_0^2$.

For the body to possess a kinetic energy equal to its initial value at a later time $t > 0$, its final angular velocity must be $\omega = -\omega_0$.

Given $\omega_0 = 20 \text{ rad/s}$ and angular deceleration $\alpha = -2 \text{ rad/s}^2$, we have $\omega = -20 \text{ rad/s}$.

Using the first equation of rotational kinematics:

$\omega = \omega_0 + \alpha t$

$-20 = 20 - 2t$

$2t = 40$

$t = 20 \text{ s}$