The radius of gyration of a 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:

For a uniform rod of length $$l$$ and mass $$M$$, the standard formula for the moment of inertia about an axis perpendicular to the rod and passing through its centre (the midpoint) is stated first: $$I_{\text{centre}}=\dfrac{1}{12}\,M\,l^{2}.$$

Now the required axis is still perpendicular to the rod but is displaced along the length by a distance $$d=\dfrac{l}{4}$$ from the centre. To shift the axis, we invoke the Parallel-Axis Theorem, which states: $$I = I_{\text{centre}} + M\,d^{2}.$$

Substituting the known values, we have $$I = \dfrac{1}{12}\,M\,l^{2} + M\left(\dfrac{l}{4}\right)^{2}.$$

Evaluating the square, $$\left(\dfrac{l}{4}\right)^{2}=\dfrac{l^{2}}{16},$$ so $$I = \dfrac{1}{12}\,M\,l^{2} + M\left(\dfrac{l^{2}}{16}\right).$$

Collecting the terms with a common denominator of $$48$$, $$I = M\,l^{2}\left(\dfrac{4}{48}+\dfrac{3}{48}\right)=M\,l^{2}\left(\dfrac{7}{48}\right).$$

The radius of gyration $$k$$ is defined through $$I = M\,k^{2}.$$ Setting $$I = M\,k^{2}$$ and cancelling the common factor $$M$$, we obtain $$k^{2}=l^{2}\left(\dfrac{7}{48}\right).$$

Taking the positive square root, $$k = l\sqrt{\dfrac{7}{48}}.$$

Hence, the correct answer is Option C.