To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.
In the case of work, we know that it is defined by,

Where,
F= Force
d = Distance
The distance in this case is a composition between number of steps and the height. Then,
, for h as the height of each step and N number of steps.
On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

Where,
Final velocity
Initial Velocity
a = Acceleration
Displacement
PART A) For the particular case of work we know then that,




Therefore the Work to do that activity is 4.41kJ
PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

Here,
3 steps in one second

Replacing,


Re-arrange for a,


At this point we can calculate the time, which is,



With time and work we can finally calculate the power
P = \frac{W}{t} = \frac{4.41}{20}
P = 0.2205kW