\(\Rightarrow \frac{\Delta V}{V}=3 \frac{\Delta R}{R}\)

\(\Rightarrow \frac{\Delta R}{R}=0.33\)

As there is no external torque angular momentum is conserved, \(L=I \omega\)

\(\Rightarrow L=\frac{2}{5} M R^{2} \omega\)

\(\Rightarrow \frac{\Delta L}{L}=2 \frac{\Delta R}{R}+\frac{\Delta \omega}{\omega}\)

since, \(\frac{\Delta L}{L}=0\) as the magnitude is constant.

\(\Rightarrow \frac{\Delta \omega}{\omega}=-2 \frac{\Delta R}{R}\)

Therefore, \(\omega\) decreases by \(0.67 \%\)