The radius of gyration of a solid sphere of radius ‘R’ about its tangential axis is

The parallel axis theorem provides the spherical shell's moment of inertia about the tangential axis as

$I_{axis}=M{k}^2$

For solid sphere about tangent,

$$\begin{gathered} I=\frac{2}{5}M{R}^2+M{R}^2=\frac{7}{5}M{R}^2 \\ \frac{7}{5}M{R}^2=M{k}^2 \\ k=\sqrt{\frac{7}{5}}R \end{gathered}$$

Hence the correct answer is $\sqrt{\frac{7}{5}}R.$