Q.At what distance below the surface of the Earth does the acceleration due to gravity decrease by 10% of its value at the surface, given radius of Earth is 6400 km.

Modelling a planet as built of many thin, uniform concentric spherical shells, the shell theorem shows that at a depth d below the surface, only the mass of the smaller INNER sphere of radius $(R-d)$ contributes any net gravitational pull -- the outer shell (of thickness d) surrounding that point contributes exactly zero, since the point lies inside it. Assuming uniform density, this leads to the strikingly simple, EXACTLY linear relation $g_d=g\left(1-\dfrac{d}{R}\right)$, valid all the way from the surface ($d=0$, $g_d=g$) down to the very centre ($d=R$, $g_d=0$).

The result that g is exactly zero at a planet's centre is a direct and important consequence of the shell theorem's cancellation, not a special assumption -- a body at the centre is pulled equally in every direction by the surrounding mass, and all these pulls cancel by symmetry, exactly as for a point anywhere inside a uniform hollow shell. Combining the depth relation (linear, for $r<R$) with the altitude relation (inverse-square, for $r>R$) into a single graph of g versus distance from the centre shows both pieces meeting at exactly $r=R$, the point of MAXIMUM gravity, with g falling off on either side whether one goes up or down from the surface.