A sphere of relative density $\sigma$ and diameter D has concentric cavity of diameter d. It will just float on water in a tank if the ratio D/d is

Let $\rho$ be the density of the sphere and ${\rho }_0$ that of water. Volume of metal =$\frac{4}{3}\mathrm{\pi }\left({\mathrm{R}}^3-{\mathrm{r}}^3\right)$ and volume of water displaced =$\frac{4}{3}\pi R^3$. From the principle of floatation, we have 

$$\begin{gathered} \frac{4}{3}\mathrm{\pi }\left({\mathrm{R}}^3-{\mathrm{r}}^3\right)\mathrm{\rho g}=\frac{4}{3}{\mathrm{\pi R}}^3{\mathrm{\rho }}_0\mathrm{g} \\ \mathrm{or} \left({\mathrm{R}}^3-{\mathrm{r}}^3\right)\frac{\mathrm{\rho }}{{\mathrm{\rho }}_0} = {\mathrm{R}}^3 \\ \mathrm{or} \left({\mathrm{R}}^3-{\mathrm{r}}^3\right)\mathrm{\sigma }={\mathrm{R}}^3 \\ \mathrm{which} \mathrm{gives} \frac{\mathrm{R}}{\mathrm{r}}={\left(\frac{\mathrm{\sigma }}{\mathrm{\sigma }-1}\right)}^{1/3} \end{gathered}$$

Hence the correct choice is (c).