@Rocca I would guess, and it is only a guess, that they mean probability density functions have to integrate to one, because the total probability must be one. So it means choose the value of $c$ such that $\int_{-\infty}^\infty f(x) dx = 1$
Now we need to know the potential change when we go from $r=2a$ to $r=a$.
Inside the larger shell the field due to its charge is zero, so between $r=2a$ and $r=a$ the only field is that due to the smaller shell. So the potential change is just:
Remember that potentials add, so we can do this problem by calculating the potential change if only the large sphere was present, then by calculating the potential if only the small sphere was present, then just adding the two.
Now we move through the outer sphere so we are inside it. And we are assuming the sphere is very thin so there is no change in potential as we move through it. So at the inside surface of the larger sphere the potential is still $U = -k 3Q/2a$. Yes?
Now the question is how much does the potential change by as we move from the inner surface of the large sphere at $r=2a$ to the outer surface of the small sphere at $r=a$?
And we can find this because inside the large sphere the field due to the charge on the large sphere is zero, so the field is only due to the charge on the small sphere. Yes?
Yes, because inside the larger sphere the field from the charge on the larger sphere is zero so it has no effect. The change in potential is just due to the +Q charge on the smaller sphere.
And we are outside the middle sphere, so the potential due to the middle sphere is $U_2 = +2kQ/r$, where $2a < r < 3a$ (you didn't say what value of $r$ you wanted).