$\theta=5 t^{2}-8 t$ $$ \begin{aligned} & \omega=\frac{\mathrm{d} \theta}{\mathrm{dt}}=10 \mathrm{t}-8 \\ & \alpha=\frac{\mathrm{d} \omega}{\mathrm{dt}}=10 \\ & \therefore \mathrm{p}=\tau \omega \\ & =(\mathrm{I} \alpha) \omega \\ & =\left(\frac{\mathrm{mR}^{2}}{2}\right) \alpha \omega \\ & =\left(\frac{\mathrm{mR}^{2}}{2}\right)(10)(10 \mathrm{t}-8) \end{aligned} $$ Put $\mathrm{t}=2$
$\mathrm{p}=60 \mathrm{mR}^{2}$