The Disc is retarded to 0 rad/s

The first assumption tells us that the disc does not translate on either the wall or floor. We can hence consider translational equilibrium first.

$\displaystyle \sum F_x = \sum F_y = 0$

Here I am denoting the frictional force and the normal contact force exerted by the wall as $f_1$ and $N_1$ respectively , frictional force and the normal contact force exerted by the floor as $f_2$ and $N_2$ respectively.

$\displaystyle \sum F_x = N_1 - f_2 = 0$

$\displaystyle \sum F_y = f_1 + N_2 - mg = 0$

$f_1 = \mu N_1 \text{ and } f_2 = \mu N_2$

Analyzing the rotational motion ,

The frictional forces $f_1 \text{ and } f_2$ will provide retarding torque, and hence angular retardation (say $\alpha$ ) about the center of the disc.

Writing torque about the center,

$f_1 R + f_2 R = \dfrac{mR^2}{2} \alpha$

To find time taken to retard to $0 rad/s$ ,

$0 - \omega_0 = - \alpha t \Rightarrow t = \dfrac{\omega_0}{\alpha}$

Solving these equations, we get

$f_1 = \frac{\mu^2 mg}{\mu^2 + 1}$

$f_2 = \frac{\mu mg}{\mu^2 + 1}$

$\alpha = \dfrac{2(\mu^2 + \mu)g}{(\mu^2 + 1)R}$

$t = \dfrac{\omega_0 (\mu^2 + 1)R}{2(\mu^2 + \mu)g} = \boxed{3}$