Explanation

(c) Escape velocity of the planet is $v_{p} = \sqrt{\frac{2GM_{P}}{R_{P}}}$
Where $M_{p}$ and $R_{p}$ be the mass and radius of the planet respectively
Escape velocity of the earth is $v_{e} = \sqrt{\frac{2GM_{e}}{R_{e}}}$
Where $M_{e}$ and $R_{e}$ be the mass and radius of the earth respectively
According to given problem, $v_{p} = 3v_{e}$ and $R_{p} = 4R_{e}$
$\therefore\ \sqrt{\frac{2GM_{p}}{4R_{e}}} = 3\sqrt{\frac{2GM_{e}}{R_{e}}} \Rightarrow \frac{M_{p}}{4R_{e}} = \frac{9M_{e}}{R_{e}}$
$\Rightarrow M_{p} = 36M_{e} = 36 \times 6 \times 10^{24}kg$
$= 216 \times 10^{24}kg = 2.16 \times 10^{26}kg$