Let us write the expression for the acceleration due to gravity at the surface of the earth in terms of density of the earth,

we can write: $\mathrm{mg}=\frac{\mathrm{GMm}}{{\mathrm{R}}^2}\Rightarrow \mathrm{g}=\frac{\mathrm{G}\left(4/3{\mathrm{\pi R}}^3\mathrm{\rho }\right)}{{\mathrm{R}}^2}$

$\mathrm{g}=\frac{4}{3}\mathrm{\pi GR\rho }$   ………(i)

Suppose that the radius of the earth has to be increased to R' by putting up more soil around it, so that the value of acceleration due to gravity will become ${\mathrm{g}}^{\mathrm{\prime }}=10{\mathrm{ms}}^{-2}$ exactly.

Then, ${\mathrm{g}}^{\mathrm{\prime }}=\frac{4}{3}{\mathrm{\pi GR}}^{\mathrm{\prime }}\mathrm{\rho }$

From Eqs. (i) and (ii), we get $\frac{{\mathrm{g}}^{\mathrm{\prime }}}{\mathrm{g}}=\frac{{\mathrm{R}}^{\mathrm{\prime }}}{\mathrm{R}}$

or ${\mathrm{R}}^{\mathrm{\prime }}=\frac{{\mathrm{g}}^{\mathrm{\prime }}}{\mathrm{g}}\mathrm{R}=\frac{10}{9.8}\times 6.4\times {10}^6=6.53\times {10}^6\mathrm{m}$

Therefore, thickness of the required layer of the soil,

$\mathrm{t}={\mathrm{R}}^{\mathrm{\prime }}-\mathrm{R}=6.53\times {10}^6-6.4\times {10}^6=0.13\times {10}^6\mathrm{m}=130\mathrm{km}$