Given the initial speed, $180 \mathrm{rpm}=180\times \frac{2\mathrm{\pi }}{60} \mathrm{rad} {\mathrm{s}}^{-1}$and final speed of the flywheel $360 \mathrm{rpm}=360\times \frac{2\mathrm{\pi }}{60} \mathrm{rad} {\mathrm{s}}^{-1}$

The energy given to the flywheel will increase the rotational kinetic energy of the flywheel, therefore, change in rotational kinetic energy will be $∆KE=\frac{1}{2}I{\left(∆\omega \right)}^2=684 \mathrm{J}$

$$\begin{gathered} 684 \mathrm{J}=\frac{1}{2}I{\left({\omega }_f-{\omega }_i\right)}^2 \\ \Rightarrow I=\frac{684\times 2}{{\left(360\times {\displaystyle \frac{2\mathrm{\pi }}{60}}-180\times \frac{2\mathrm{\pi }}{60}\right)}^2}=\frac{1368}{36{\mathrm{\pi }}^2}=3.85 \mathrm{kg} {\mathrm{m}}^2 \end{gathered}$$