The acceleration due to gravity at a radial distance r (r < R) from the centre of a sphere of constant mass density $$\rho$$ is given by

$$g = {{G\left( {{4 \over 3}\pi {r^3}} \right)\rho } \over {{r^2}}} = {{4\pi \rho G} \over 3}r$$.

Consider a spherical shell of radius r and thickness dr.

Let P and P + dP be the fluid pressures inside and outside the shell. Consider a small element of area A and mass $$dm = \rho Adr$$ on the shell. The forces on this element are gravitational force (dm g) direted inwards, the force due to pressure of the fluid outside the shell $$((P + dP)A)$$ directed inwards, and the force due to pressure of the fluid inside the shell (PA) directed outwards. In equilibrium, net force on the element is zero, i.e.,

$$(\rho Adr){4 \over 3}\pi \rho Gr + (P + dP)A = PA$$, or

$$dP = - {4 \over 3}\pi G{\rho ^2}r\,dr$$.

Note that the pressure decreases with increase in r. The pressure is zero at the surface of the sphere i.e., P = 0 at r = R. Integrate to get

$$\int_0^{P(r)} {dP = P(r) = - {4 \over 3}\pi G{\rho ^2}\int_R^r {rdr} } $$

$$ = {2 \over 3}\pi G{\rho ^2}({R^2} - {r^2})$$.

Now, $$P(r = 0) = {{2\pi } \over 3}{\rho ^2}G{R^2} \ne 0$$

$${{P(r = 3R/4)} \over {P(r = 2R/3)}} = {{{R^2} - {{(3R/4)}^2}} \over {{R^2} - {{(2R/3)}^2}}} = {7 \over {16}} \times {9 \over 5} = {{63} \over {80}}$$

$${{P(r = 3R/5)} \over {P(r = 2R/5)}} = {{{R^2} - {{(3R/5)}^2}} \over {{R^2} - {{(2R/5)}^2}}} = {{16} \over {25}} \times {{25} \over {21}} = {{16} \over {21}}$$

$${{P(r = R/2)} \over {P(r = R/3)}} = {{{R^2} - {{(R/2)}^2}} \over {{R^2} - {{(R/3)}^2}}} = {3 \over 4} \times {9 \over 8} = {{27} \over {32}} \ne {{20} \over {27}}$$