Let the density of earth is $d$ and radius $R$

Then the density of the planet will be $8 \mathrm{d}$ and radius $2 \mathrm{R}$

Mass of Earth

$M=\frac{4}{3} \pi R^{3} \times d$

Mass of the planet

$M^{\prime}=\frac{4}{3} \pi \times(2 R)^{3} \times 8 d$

$\Longrightarrow M^{\prime}=64 M$

On Earth the acceleration due to gravity

$g=\frac{G M}{R^{2}}$

On the planet, the acceleration due to gravity of the planet

$g^{\prime}=\frac{G M^{\prime}}{(2 R)^{2}}$

$\Longrightarrow g^{\prime}=\frac{G \times 61 M}{4 R^{2}}$

$\Longrightarrow g^{\prime}=16 g$

On earth a mass of $2 \mathrm{kg}$ travels distance $\mathrm{S}$ in $1$ second

Therefore, using the second equation of motion

$S=0 \times 1+\frac{1}{2} g \times 1^{2}$

$\Longrightarrow S=\frac{g}{2}$

$\Longrightarrow 2 S=g$

Now on the planet if the time taken by the mass of $4 \mathrm{kg}$ to fall is $t$ then the acceleration due to the gravity of the planet acting on the body will be $g'$

Note that the acceleration due to the planets in either case will not depend upon the mass of the objects

Thus,

$S=0 \times t+\frac{1}{2} g^{\prime} t^{2}$

$\Longrightarrow S=\frac{1}{2} \times 16 g \times t^{2}$

or, $S=8 g t^{2}$

or, $S=8 \times 2 S \times t^{2}$

or, $t^{2}=\frac{1}{16}$

or, $t=\sqrt{\frac{1}{16}}$

or, $t=\frac{1}{4}$ seconds

or, $t=0.25$ seconds