Gauss' Law

The fact that $E=Ae^{-br}/r$ makes the problem quite easy,as the field behaves like $1/r$ in the limit $r$ tending to zero there isn't the case of a Dirac Function.So we simply use the Divergence theorem, $\int _{V}(\nabla \cdot E)d\tau =\int _{S} E\cdot da$ and take the limit of radius tending to infinite to get the total charge. To state a better version of the same problem consider a filed where the potential varies as $V(r)=A\frac {e^{-br}}{r}$ then $E=-\frac {dV}{dr}=\frac {e^{-br}(1+br)}{r^2}$ ,clearly this behaves as a $1/r^2$ field in the limit of $r$ tending to zero.Hence we need the Dirac delta function.From Gauss law $\rho =\epsilon _{o}\nabla .E$ hence we get $\rho _{r}=A\epsilon _{o}(4\pi \delta ^3(r)-\frac {b^2e^{-br}}{r})$ ,using the fact $\int _{all space}\delta ^3(r)d\tau =1$ we get $Q=\int \rho d\tau =0$

I don't know but Dirac delta but simple gauss law and some integration does the job.