Answer:
The escape velocity on the planet is approximately 178.976 km/s
Explanation:
The escape velocity for Earth is therefore given as follows
The formula for escape velocity,
, for the planet is 
Where;
= The escape velocity on the planet
G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
m = The mass of the planet = 12 × The mass of Earth, 
r = The radius of the planet = 3 × The radius of Earth, 
The escape velocity for Earth,
, is therefore given as follows;


= 16 ×
Given that the escape velocity for Earth,
≈ 11,186 m/s, we have;
The escape velocity on the planet =
≈ 16 × 11,186 ≈ 178976 m/s ≈ 178.976 km/s.