A thin uniform rod of length 2   m , cross sectional area A and density d is rotated about an axis…

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

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

$\Rightarrow E=\frac{1}{2}\frac{dA{l}^3}{12}{\omega }^2 \left(\because \mathrm{mass}=\mathrm{density}\times \mathrm{volume}\right)$

$\Rightarrow E=\frac{dA(2{)}^3}{24}{\omega }^2$

$\Rightarrow \omega =\sqrt{\frac{3E}{dA}}$

Thus, the value of $\alpha =3$.