Acceleration due to gravity or the strength of gravitational field is obtained from Gauss's law :

$g \times 4πr^2=-4πG\displaystyle \int_0^r kr^n\times 4πr^2dr$ (here $G$ is the Universal Gravitational Constant)

$\implies g=-πkGr^{n+1}, g_0=-πkGR^{n+1}\implies$

$g=g_0(\frac{r}{R})^{n+1}$ .

In this problem, $g_0=63, r=9000-3000=6000, R=9000, n=1$ , in appropriate units, and so $g=63(\frac{6000}{9000})^2=\boxed {28}$ m/s $^2$ .

Calculating mass m of planet at a distance r from center

$dm = d(\rho V)$

$dm = (kx)(4πx^2)dx$

$\int_0^mdm = 4\pi k\int_0^r{x^3}dx$

$m = kπr^4$

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Now, calculating acceleration due to gravity a at a distance r from center

$a = \frac{Gm}{r^2}$

$a = \frac{Gkπr^4}{r^2}$

$a = Gkπr^2$

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At r = R, a = g and at r = R', g = g'

$g = GkπR^2 \text{and } g' = GkπR'^2$

$\frac{g'}{g} = \frac{GkπR'^2}{GkπR^2}$

$g' = g\frac{R'^2}{R^2}$

At depth d = R - R'

$\boxed{g' = g\bigg(1 - \frac{d}{R}\bigg)^2}$

$g' = (63)\bigg(1 - \frac{3000}{9000}\bigg)^2$

$\color{#3D99F6}{\boxed{g = 28m/s^2}}$