The radius of gyration of a cylindrical rod about an axis of rotation perpendicular to its length and…

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

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

$$\begin{gathered} \Rightarrow k^2=\frac{l^2}{12} \\ \Rightarrow k=\frac{l}{\sqrt{12}}=\frac{l}{2\sqrt{3}}=\frac{10\sqrt{3}}{2\sqrt{3}}=5 \mathrm{m} \end{gathered}$$