I'm not that smart but I think it is c I really hope It helps
If the distance between two charges is halved, the electrical force between them increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:

where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:

the magnitude of the force changes as follows:

so, the force increases by a factor 4.
Answer:
Slope = 2 m / 10 m = 1/5
For every 5 m of effort the object will be raised 1 m
W = work done on object = M g h increase in PE of object
E S = W where E is effort and S the distance thru which the effort acts
E S = M g H
E = 100 kg * 9.8 m/s^2 * 2 m / 10 m = 196 kg m / s^2 = 196 N
Check: total work = 2 * 9.8 * 100 = 1960 J
Force Needed = 1960 J / 2 m = 980 Newtons
Mechanical advantage = 980 / 196 = 5 as one would expect since the object is raised 1 m for every 5 m of force input