#### Answer

The brake must apply a friction force of 1.6 N

#### Work Step by Step

We can find the moment of inertia of the disk.
$I = \frac{1}{2}MR^2$
$I = \frac{1}{2}(2.0~kg)(0.15~m)^2$
$I = 0.0225~kg~m^2$
We can convert the initial angular velocity to units of rad/s
$\omega_0 = (300~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$
$\omega_0 = 31.42~rad/s$
We can find the required angular acceleration for the disk to stop in 3.0 seconds.
$\alpha = \frac{\omega_f-\omega_0}{t}$
$\alpha = \frac{0-31.42~rad/s}{3.0~s}$
$\alpha = -10.47~rad/s^2$
We can find use the magnitude of the angular acceleration to find the required force of friction.
$\tau = I~\alpha$
$(R)~(F_f) = I~\alpha$
$F_f = \frac{I~\alpha}{R}$
$F_f = \frac{(0.0225~kg~m^2)(10.47~rad/s^2)}{0.15~m}$
$F_f = 1.6~N$
The brake must apply a friction force of 1.6 N