The mass density of a planet of radius R varies with the distance r from its centre as $$\rho(r) = \rho_0\left(1 - \frac{r^2}{R^2}\right)$$. Then the gravitational field is maximum at:

We are told that the mass-density inside the planet depends on the distance from the centre according to

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

To find the point where the gravitational field $$g(r)$$ is maximum we must first obtain an explicit expression for $$g(r)$$ inside the planet and then differentiate it with respect to $$r$$.

The gravitational field at a point situated at a distance $$r<R$$ from the centre depends only on the mass contained within the sphere of radius $$r$$. By Gauss’s law for gravitation (equivalent to Newton’s law), we have

$$g(r)=\frac{G\,M(r)}{r^{2}},$$

where $$M(r)$$ is the mass enclosed within radius $$r$$.

To calculate $$M(r)$$, we integrate the density over the volume from the centre up to $$r$$:

$$M(r)=\int_{0}^{r} 4\pi r'^{2}\,\rho(r')\,dr'.$$

Substituting $$\rho(r')=\rho_0\!\left(1-\dfrac{r'^{2}}{R^{2}}\right)$$, we get

$$M(r)=4\pi\rho_0\!\int_{0}^{r}\!\left(1-\frac{r'^{2}}{R^{2}}\right)r'^{2}dr'.$$

Expanding inside the integral,

$$M(r)=4\pi\rho_0\!\left[\int_{0}^{r}r'^{2}dr'-\frac{1}{R^{2}}\int_{0}^{r}r'^{4}dr'\right].$$

Using the power-integral formulas $$\displaystyle\int r'^{2}dr'=\frac{r'^{3}}{3}$$ and $$\displaystyle\int r'^{4}dr'=\frac{r'^{5}}{5},$$ we obtain

$$M(r)=4\pi\rho_0\!\left[\frac{r^{3}}{3}-\frac{r^{5}}{5R^{2}}\right].$$

Now we substitute this result into $$g(r)=\dfrac{G\,M(r)}{r^{2}}$$:

$$g(r)=\frac{G}{r^{2}}\;4\pi\rho_0\!\left[\frac{r^{3}}{3}-\frac{r^{5}}{5R^{2}}\right].$$

Simplifying the factors of $$r$$,

$$g(r)=4\pi G\rho_0\left[\frac{r}{3}-\frac{r^{3}}{5R^{2}}\right].$$

It is convenient to pull out a common constant $$k=4\pi G\rho_0$$, so

$$g(r)=k\,r\!\left(\frac13-\frac{r^{2}}{5R^{2}}\right).$$

To locate the maximum of $$g(r)$$ for $$0<r<R$$ we differentiate with respect to $$r$$ and set the derivative to zero. Let

$$f(r)=r\!\left(\frac13-\frac{r^{2}}{5R^{2}}\right) \quad\text{so that}\quad g(r)=k\,f(r).$$

Differentiating term by term, we have

$$\frac{df}{dr}=\left(\frac13-\frac{r^{2}}{5R^{2}}\right)+r\!\left(-\frac{2r}{5R^{2}}\right).$$

The two terms give

$$\frac{df}{dr}=\frac13-\frac{r^{2}}{5R^{2}}-\frac{2r^{2}}{5R^{2}}=\frac13-\frac{3r^{2}}{5R^{2}}.$$

Setting $$\dfrac{df}{dr}=0$$ for an extremum,

$$\frac13-\frac{3r^{2}}{5R^{2}}=0.$$ $$\frac{3r^{2}}{5R^{2}}=\frac13.$$ $$r^{2}=R^{2}\left(\frac13\cdot\frac{5}{3}\right)=\frac{5}{9}R^{2}.$$

Taking the positive square root (since $$r$$ is a radial distance),

$$r=\sqrt{\frac{5}{9}}\;R=\frac{\sqrt5}{3}\,R.$$

This value indeed lies within the planet ($$r<R$$) and represents the point where $$g(r)$$ attains its maximum. Comparing with the given choices, this corresponds to Option D.

Hence, the correct answer is Option D.