A non – homogeneous sphere of radius R has the following density variation

$\rho ={\rho }_{0}forr\le R/3$

$\rho =\frac{{\rho }_0}{2}for\frac{R}{3}<r\le \frac{3R}{4}and$

$\rho =\frac{{\rho }_0}{8}for\frac{3R}{4}<r\le R$

The gravitational field due to the sphere at $r=\frac{5R}{6}$ is

For $r=\frac{5R}{6},upto\frac{R}{3},\rho ={\rho }_0,$ between $\frac{R}{3}and\frac{3R}{4},$

$\rho =\frac{{\rho }_0}{2}$ and between $\frac{3R}{4}$ and $\frac{5R}{6},$ $\rho =\frac{{\rho }_0}{8}$

$\therefore$  The mass of the sphere of radius 5R/6 is

$m=\frac{4}{3}\pi {\left(\frac{R}{3}\right)}^3{\rho }_0+\left[\frac{4}{3}\pi {\left(\frac{3R}{4}\right)}^3-\frac{4}{3}\pi {\left(\frac{R}{3}\right)}^3\right]{\rho }_0/2$

$+\left[\frac{4}{3}\pi {\left(\frac{5R}{6}\right)}^3-\frac{4}{3}\pi {\left(\frac{3R}{4}\right)}^3\right]{\rho }_0/8$

$=0.332\pi R^3{\rho }_0$

$\therefore$  Gravitational field at a distance, $r=\frac{5R}{6}$ is

$E_g=\frac{Gm}{{\left(5R/6\right)}^2}$

$=\frac{G\times 0.332\pi R^3{\rho }_0}{25R^2}\times 36$

$=0.48\pi GR{\rho }_0$