If the density of a small planet is the same as that of earth, while the radius of the planet is $0.2$ times that the earth, the gravitational acceleration on the surface of that planet is:

Acceleration due to gravity               $g=\frac{GM}{R^2}$

Mass of the planet,       $M=Vd=d\times \frac{4}{3}\pi R^3$          where  $d$ is the density of the planet

$⟹$      $g=\frac{4}{3}\pi dGR$              (for earth)

Given :          $d^{\prime }=d$   and    $R^{\prime }=0.2R$

$\therefore$        $g^{\prime }=\frac{4}{3}\pi d^{\prime }G{R}^{\prime }=\frac{4}{3}\pi dG(0.2R)$

$⟹$    $g^{\prime }=0.2g$