#### Answer

A torque of $0.28~N~m$ will bring the balls to a halt in 5.0 seconds.

#### Work Step by Step

We first express the initial angular velocity in units of rad/s. Note that since the rotation is clockwise, the angular velocity is negative.
$\omega_0 = -(20~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$
$\omega_0 = -2.094~rad/s$
We then find the angular acceleration if the object comes to rest in 5.0 seconds.
$\alpha = \frac{\omega_f-\omega_0}{t}$
$\alpha = \frac{0-(-2.094~rad/s)}{5.0~s}$
$\alpha = 0.419~rad/s^2$
Next, we find the moment of inertia of the object. Note that the center of mass is 33 cm from the 2.0-kg ball.
$I = m_1~R_1^2+m_2~R_2^2$
$I = (2.0~kg)(0.33~m)^2+(1.0~kg)(0.67~m)^2$
$I = 0.67~kg~m^2$
We then find the required torque:
$\tau = I~\alpha$
$\tau = (0.67~kg~m^2)(0.419~rad/s^2)$
$\tau = 0.28~N~m$
A torque of $0.28~N~m$ will bring the balls to a halt in 5.0 seconds.