1. Moment of inertia of a uniform rod about its centre

For a uniform rod of length $l$ and mass $m$, the moment of inertia about an axis through its centre and perpendicular to the rod is

$I_{\text{cm}} = \frac{1}{12}ml^2.$

2. Shift the axis using the parallel axis theorem

The new axis passes through a point at a distance $\dfrac{l}{4}$ from the centre of the rod, and is perpendicular to the rod.

By the parallel axis theorem,

$I = I_{\text{cm}} + md^2$

where

$d = \frac{l}{4}.$

So,

$I = \frac{1}{12}ml^2 + m\left(\frac{l}{4}\right)^2.$

3. Simplify

$I = \frac{1}{12}ml^2 + \frac{1}{16}ml^2$

Taking LCM $48$,

$I = \left(\frac{4}{48} + \frac{3}{48}\right)ml^2 = \frac{7}{48}ml^2.$

4. Use the definition of radius of gyration

If $k$ is the radius of gyration, then

$I = mk^2.$

Thus,

$mk^2 = \frac{7}{48}ml^2$

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

$k = \sqrt{\frac{7}{48}}\,l.$

5. Match with the options

$k = \sqrt{\frac{7}{48}}\,l$

This corresponds to Option C.

Verification with stored answer: Stored correct answer is C, which matches our result.