Given $: K_{ \text{rot }}=2.5\, J , \omega_{1}=\omega, \omega_{2}=10 \omega$
Rotational kinetic energy
$K_{\text {rot }}=\frac{1}{2} I \omega^{2}$
or $K_{\text{rot }} \propto \omega^{2}$
$\therefore \frac{K_{\text {rot }}'}{K_{\text {rot }}'}=\left(\frac{\omega_{2}}{\omega_{1}}\right)^{2}=(10)^{2}=100 $
$ \therefore K_{\text {rot }}'=100 \times 2.5=250\, J $