Consider two point charges that are separated by a distance,2r.if this distance between them is increased to 5r, the force between the charges is what?

Use Coloumb's formula: the force between point charges q_1,q_2 separated by a distance
r is

\frac{1}{4\pi\vep_0} \frac{q_1q_2}{r^2}

Thus changing the distance from 2r to 5r will change the force by

\frac{\frac{1}{4\pi\vep_0} \frac{q_1q_2}{(2r)^2}}{\frac{1}{4\pi\vep_0} \frac{q_1q_2}{(5r)^2}}
=25/4
thus if you call F the force between your charges with 2r between them, when the distance is 5r the force between them is \frac{4}{25}F .

\vep_0 is a constant, not influent for your question.