Explanation

(b) The relation between mass and density of earth is given form Newton’s law of gravitation, according to which
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M_{e} = \frac{{gR}_{e}^{2}}{G}$
where $M_{e}$ is mass of earth, $G$ the gravitational constant, $R_{e}$ the radius of earth and $g$ the acceleration due to gravity.
$Also,\ \ mass = volume \times density$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = \frac{G \times volume \times density}{R^{2}}$
Assuming spherical shape of earth volume
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{4}{3}{\pi R}^{3}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = G \times \frac{4}{3}\frac{{\pi R}^{3}}{R^{2}}\rho$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g = G \bullet \frac{4}{3}\pi R\rho$
Hence, increases in radius would dominate.