vysero said:
I am not sure what you mean by an induced charge distribution. I didn't get this part when my professor tried explaining it in class either. Should I assume that the net charge for the all the shells is 19-4+1 = 16uC?
Yes, that's the net charge of the system.
Because each shell is a conductor, the charges on it will be distributed on the surfaces. I.e. a charge distribution on the inside of each shell and another on the outside of each shell.
By the symmetry of the set-up, each of these six charge distributions will be uniform.
The total charge on each shell is given, so there are in effect three unknowns. The charges on each shell will arrange themselves so that there is no net field (after taking into account all the other shells) within its conducting material.
In the special case of a hollow conductor with no charged objects in the hollow, there is no field in the hollow either.
So, considering potentials, there is a uniform potential within the central hollow and continuing through the innermost shell to its outer surface. There may be a potential gradient (hence a field) from there to the inner surface of the next shell, then no further change in potential until its outer surface, and so on.
One way to attack this problem is to assign a symbolic variable to each of the six charges (or to the charge densities if you prefer). You have immediately the three equations for the total charge on each shell.
Next, you can calculate the potential at any given point in the system in terms of those six charges. (To do this you have to know the standard formulae for potential inside and outside uniformly charged spherical shells. These are essential knowledge for the subject. They should have been quoted in the OP as relevant equations.)
You can now write down three more equations corresponding to the fact that for each shell the potential is the same at its inner surface and its outer surface.
Six equations, six unknowns. Solve.
That gives you all the charge distributions. At any given point, you can compute the field due to each charge distribution and add them (vectorially)... or you can find the expression for the potential in a small neighbourhood of the point and differentiate to find the field.