The radius of gyration of uniform rod of length ‘l’ about an axis passing through a point $\frac{l}{4}$  away from the centre of the rod and perpendicular to it is

Moment of inertia of the rod passing a point $\frac{l}{4}$  away from the centre of the Rod. 

$I\mathit{=}{I}_{\mathit{c}}\mathit{+}m{L}^{\mathit{2}}\mathit{(}\mathit{L}\mathit{=}\frac{\mathit{l}}{\mathit{4}}\mathit{)}$

$=\frac{m{l}^2}{12}+m{\left(\frac{l}{4}\right)}^2$

$=\frac{m{l}^2}{12}+\frac{m{l}^2}{16}$

$=\frac{4m{l}^2+3m{l}^2}{48}$

$I=\frac{7}{48}m{l}^2$
using $I=m{k}^2$

$\begin{array}{l}\cancel{m}{k}^2=\frac{7}{48}\cancel{m}{l}^2\\ k=\sqrt{\frac{7}{48}}l\end{array}$