The radius of gyration of a cylindrical rod about an axis of...

Moment of inertia of the rod about an axis perpendicular to its length and passing through the centre is $I = \dfrac{m{l}^{2}}{12}$.

Using the relation, $I = m{k}^{2}$, we have

$⇒ {k}^{2} = \dfrac{{l}^{2}}{12} ⇒ k = \dfrac{l}{\sqrt{12}} = \dfrac{l}{2\sqrt{3}} = \dfrac{10\sqrt{3}}{2\sqrt{3}} = 5m$