A thin uniform rod of length 2m , cross sectional area A and...

Rotational kinetic energy is given by $E = \dfrac{1}{2}I{\omega}^{2}$, here, $I$ is moment of inertia of rod about an axis through its centre and perpendicular to it.

So, $E = \dfrac{1}{2}\dfrac{m{l}^{2}}{12}{\omega}^{2}$

$⇒ E = \dfrac{1}{2}\dfrac{dA{l}^{3}}{12}{\omega}^{2}\left(∵ mass = density \times volume\right)$

$⇒ E = \dfrac{dA ( 2{)}^{3}}{24}{\omega}^{2}$

$⇒ \omega = \sqrt{\dfrac{3E}{dA}}$

Thus, the value of $\alpha = 3$.