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This is right out of Griffiths (problem 2.29) I have a solution manual, but I'd like to try to get a nudge in the right direction before I turn to it.

Poisson's eq:

[itex]\nabla ^2 V = - \frac{ \rho }{\epsilon }[/itex]

Using an identity out of Griffiths we have [itex]\nabla ^2\left(\frac{1}{r}\right) = -4\pi \delta ^3(r)[/itex]

finally we know that [itex] V(r) = \frac{1}{4\pi \epsilon }\int \frac{p(r')}{r} \, dr'[/itex]

distributing the gradient into our function will yield

[itex]V(r) = -\frac{1}{\epsilon }\int p(r')\delta ^3(r)dr'[/itex]

But I do not know what to do with the dirac delta in the final function we have arrived at.