A sphere is performing pure rolling as shown in figure. The radius of curvature of trajectory of point A in the shown portion is (OA = R/2)

 

From the figure,

$$\begin{gathered} {\overrightarrow{\mathrm{V}}}_{\mathrm{A}}={\overrightarrow{\mathrm{V}}}_{\mathrm{O}}+{\overrightarrow{\mathrm{V}}}_{\mathrm{A}-\mathrm{O}} \\ \therefore {\mathrm{V}}_{\mathrm{A}}=\mathrm{V}+\mathrm{\omega }\left(\mathrm{OA}\right)=\mathrm{V}+\left(\frac{\mathrm{V}}{\mathrm{R}}\right)\frac{\mathrm{R}}{2}=\frac{3\mathrm{V}}{2} \end{gathered}$$

Acceleration of A relative to the round,

$=\frac{(3\mathrm{V}/2{)}^2}{\mathrm{r}}=\frac{9{\mathrm{V}}^2}{4\mathrm{r}},$ (If r is the radius of curvature)

$$\begin{gathered} {\overrightarrow{\mathrm{a}}}_{\mathrm{A}}={\overrightarrow{\mathrm{a}}}_{\mathrm{O}}+{\overrightarrow{\mathrm{a}}}_{\mathrm{A}-\mathrm{O}} \\ \Rightarrow {\mathrm{a}}_{\mathrm{A}}=0+{\mathrm{\omega }}^2(\mathrm{R}/2)=\frac{{\mathrm{V}}^2}{2\mathrm{R}} \end{gathered}$$

$\therefore \frac{9{\mathrm{V}}^2}{4\mathrm{r}}=\frac{{\mathrm{V}}^2}{2\mathrm{R}}\Rightarrow \mathrm{r}=\frac{9\mathrm{R}}{2}$

Hence the correct answer is $\frac{9}{2}R.$