1. Condition for minimum liquid density for floating

For a body to just float in a liquid, the minimum liquid density required is equal to the average density of the body:

$\rho_{\text{liquid,min}} = \rho_{\text{avg}} = \frac{M}{V}$

So we first find the mass of the non-uniform solid sphere.

2. Given density distribution

$\rho(r)=\rho_0\left(1-\frac{r^2}{R^2}\right)$

For a thin spherical shell of radius $r$ and thickness $dr$,

$dV = 4\pi r^2 dr$

Hence,

$dM = \rho(r)\, dV = \rho_0\left(1-\frac{r^2}{R^2}\right)4\pi r^2 dr$

So total mass is

$M=\int_0^R 4\pi \rho_0\left(1-\frac{r^2}{R^2}\right) r^2 \, dr$

3. Evaluate the integral

$M=4\pi \rho_0 \int_0^R \left(r^2-\frac{r^4}{R^2}\right)dr$

Now,

$\int_0^R r^2 dr = \frac{R^3}{3}, \qquad \int_0^R r^4 dr = \frac{R^5}{5}$

Therefore,

$M=4\pi \rho_0\left(\frac{R^3}{3}-\frac{1}{R^2}\cdot \frac{R^5}{5}\right)$

$M=4\pi \rho_0 R^3\left(\frac{1}{3}-\frac{1}{5}\right)$

$M=4\pi \rho_0 R^3\left(\frac{2}{15}\right)$

$M=\frac{8\pi \rho_0 R^3}{15}$

4. Volume of the sphere

$V=\frac{4}{3}\pi R^3$

So average density is

$\rho_{\text{avg}}=\frac{M}{V} = \frac{\frac{8\pi \rho_0 R^3}{15}}{\frac{4}{3}\pi R^3}$

Canceling common factors,

$\rho_{\text{avg}}=\frac{8}{15}\cdot \frac{3}{4}\rho_0$

$\rho_{\text{avg}}=\frac{2\rho_0}{5}$

5. Final answer

The minimum density of the liquid in which the sphere will float is

$\boxed{\frac{2\rho_0}{5}}$

So the correct option is B.

6. Comparison with stored correct answer

Stored correct answer: B

Derived answer: B

They match.