A uniform thin rod of length L and a uniform solid sphere of radius...

Moment of inertia of a uniform thin rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is $I_{\text{rod}} = \dfrac{ML^2}{12}$.

The radius of gyration $k_{\text{rod}}$ is given by $k_{\text{rod}}^2 = \dfrac{L^2}{12}$.

Moment of inertia of a uniform solid sphere of mass $M$ and radius $R$ about its diametric axis is $I_{\text{sphere}} = \dfrac{2}{5}MR^2$.

The radius of gyration $k_{\text{sphere}}$ is given by $k_{\text{sphere}}^2 = \dfrac{2}{5}R^2$.

Given that $k_{\text{rod}} = k_{\text{sphere}}$, we have:

$\dfrac{L^2}{12} = \dfrac{2}{5}R^2$

$\dfrac{L^2}{R^2} = \dfrac{24}{5}$

$\dfrac{L}{R} = \sqrt{\dfrac{24}{5}}$

Answer: $\sqrt{\dfrac{24}{5}}$