Acccording to work energy theorem,

W = ${1 \over 2}\left( {\omega _f^2 - \omega _i^2} \right)$

Given that,

$\theta$ = 2$\pi$ revolution/minute

$\theta$ = 2$\pi$ $\times$ 2$\pi$ = 4$\pi$² rad

${\omega _i} = 3 \times {{2\pi } \over {60}}$ rad/s

${\omega _f} = 0$ rad/s

By putting the value of ${\omega _f}$ and ${\omega _i}$, we get

- \tau $$\theta = ${1 \over 2} \times {1 \over 2}m{r^2}\left( {0 - \omega _i^2} \right)$

$\Rightarrow$ $\tau$ = $- {{{1 \over 2} \times {1 \over 2} \times 2 \times \left( {4 \times {{10}^{ - 2}}} \right) \times {{\left( { - 3 \times {{2\pi } \over {60}}} \right)}^2}} \over {4{\pi ^2}}}$

= 2 $\times$ 10^-6 Nm