Retarding torque is constant, therefore retardation $\alpha$ will also be constant.

${\mathrm{\omega }}^2={{\mathrm{\omega }}_{\mathrm{o}}}^2-2\mathrm{\alpha \theta }$

$$\begin{gathered} \Rightarrow {\left(\frac{{\mathrm{\omega }}_{\mathrm{o}}}{2}\right)}^2={{\mathrm{\omega }}_{\mathrm{o}}}^2-2\mathrm{\alpha \theta \_\_\_\_\_\_}\left(1\right) \\ \mathrm{and} 0={\left(\frac{{\mathrm{\omega }}_{\mathrm{o}}}{2}\right)}^2-2{\mathrm{\alpha \theta }}_2\mathrm{\_\_\_\_\_\_}\left(2\right) \end{gathered}$$

solving equation 1 and 2

${\mathrm{\theta }}_2=\frac{{\mathrm{\theta }}_1}{3}$

Thus disk will make n/3 more rotation before coming to rest.

Hence the correct answer is $\frac{\mathrm{n}}{3}.$