${\mathrm{\omega }}_1=20\mathrm{ }\mathrm{rad} {\mathrm{s}}^{-1}$, $\mathrm{\omega }=0$ , $\mathrm{t}=4 \mathrm{s}$ . So angular retardation $\mathrm{\alpha }=\frac{{\mathrm{\omega }}_1-{\mathrm{\omega }}_2}{\mathrm{t}}=\frac{20}{4}=5 \mathrm{rad} {\mathrm{s}}^{-2}$

Now, angular speed after $2 \mathrm{s}$

 ${\mathrm{\omega }}_2={\mathrm{\omega }}_1-\mathrm{\alpha t}=20-5\times 2=10 \mathrm{rad} {\mathrm{s}}^{-1}$ .

Work done by torque in $2 \mathrm{s}$ = loss in kinetic energy $=\frac{1}{2}\mathrm{l}\left({\mathrm{\omega }}_1^2-{\mathrm{\omega }}_1^2\right)=\frac{1}{2}\left(0.20\right)\mathrm{ }\left({\left(20\right)}^2-{\left(10\right)}^2\right)$

$=\frac{1}{2}\left(0.20\right) \left(400-100\right)$

$=\frac{1}{2}\left(0.20\right)\times \left(300\right)=\frac{1}{2}\times 60=30 \mathrm{J}$