If the system is operated on mars, through what distance would the 18.0-kg mass have to fall to give the same amount of kinetic energy to the drum?

The previous part of the exercise says:
"Engineers are designing a system by which a falling mass m imparts kinetic energy to a rotating uniform drum to which it is attached by thin, very light wire wrapped around the rim of the drum. There is no appreciable friction in the axle of the drum, and everything starts from rest. This system is being tested on Earth, but it is to be used on Mars, where the acceleration due to gravity is 3.71 m/s². In the Earth tests, when m is set to 18.0 kg and allowed to fall through 5.50 m, it gives 300.0 J of kinetic energy to the drum."

Since Kearth = Kmars, we have, for conservation of energy, that also the potential energies must be equal:
Uearth = Umars

which means:
m · gearth · hearth = m · gmars · hmars

we can solve for hmars:
hmars = (gearth / gmars) · hearth
           = (9.8 / 3.71) · 5.50
           = 14.53m

Therefore, the correct answer will be: the mass would have to fall from an height of 14.53m.