Using the known radius of the Earth and $\mathrm{g}=9.80 \mathrm{m}/{\mathrm{s}}^2$ at the Earth surface, the average density of the Earth _______ $\times {10}^3 \mathrm{kg}/{\mathrm{m}}^3$  ( Radius of the Earth =6.37$\times {10}^6 \mathrm{m}$)

The density of material in the Earth varies, but let’s adopt a simplified model in which we assume the density to be uniform throughout the Earth. The resulting density is the average density of the Earth. Assume the Earth is a perfect shpere.
We know that $\mathrm{g}=\frac{{\mathrm{GM}}_{\mathrm{E}}}{{\mathrm{R}}_{\mathrm{E}}^2}$

The mass of the earth as ${\mathrm{M}}_{\mathrm{E}}=\frac{{\mathrm{gR}}_{\mathrm{E}}^2}{\mathrm{G}}$

substitute this mass and the volume of a sphere into the definition of density

${\rho }_{\mathrm{E}}=\frac{{\mathrm{M}}_{\mathrm{E}}}{{\mathrm{V}}_{\mathrm{E}}}=\frac{{\mathrm{gR}}_{\mathrm{E}}^2/\mathrm{G}}{\frac{4}{3}\pi {\mathrm{R}}_{\mathrm{E}}^3}=\frac{3}{4}\frac{\mathrm{g}}{\pi {\mathrm{GR}}_{\mathrm{E}}}$

$=\frac{3}{4}\frac{9.80\mathrm{m}/{\mathrm{s}}^2}{\pi (6.674\times {10}^{-11}{\mathrm{N}}^2{\mathrm{m}}^2/{\mathrm{kg}}^2)(6.37\times {10}^6\mathrm{m})}$

${\rho }_{\mathrm{E}}=5.50\times {10}^3\mathrm{kg}/{\mathrm{m}}^3$