1. Interpretation of “relative density”

Let the density of water be $\rho_w$. If the material of the sphere has relative density $\sigma$, then its density is

$\rho = \sigma \rho_w.$

The outer sphere has diameter $D$ and contains a concentric cavity of diameter $d$.

2. Condition for “just floats”

If it just floats in water, that means the body is in equilibrium when fully submerged, i.e. its average density equals the density of water.

So,

$\text{mass of body} = \rho_w \times \text{outer volume}.$

3. Write the actual mass of the hollow sphere

Outer volume:

$V_D = \frac{\pi}{6}D^3$

Cavity volume:

$V_d = \frac{\pi}{6}d^3$

Hence volume of material present is

$V_{\text{mat}} = V_D - V_d = \frac{\pi}{6}(D^3-d^3).$

So mass of the body is

$m = \rho \,(V_D - V_d)= \sigma \rho_w \frac{\pi}{6}(D^3-d^3).$

4. Apply floating condition

For just floating,

$m = \rho_w V_D = \rho_w \frac{\pi}{6}D^3.$

Substitute $m$:

$\sigma \rho_w \frac{\pi}{6}(D^3-d^3)=\rho_w \frac{\pi}{6}D^3.$

Cancel common factors $\rho_w$ and $\frac{\pi}{6}$:

$\sigma(D^3-d^3)=D^3.$

Expand:

$\sigma D^3-\sigma d^3=D^3.$

$\sigma D^3-D^3=\sigma d^3.$

$D^3(\sigma-1)=\sigma d^3.$

Therefore,

$\left(\frac{D}{d}\right)^3 = \frac{\sigma}{\sigma-1}.$

Taking cube root,

$\frac{D}{d}=\left(\frac{\sigma}{\sigma-1}\right)^{1/3}.$

5. Match with options

This corresponds to:

$\boxed{\left(\frac{\sigma}{\sigma-1}\right)^{1/3}}$

So the correct option is D.

6. Comparison with stored answer

Stored correct answer: D

Derived answer: D

They agree.