I've some difficulty deriving this equation from jackson electrodynamics (The equation after 1.30)
$\nabla^2 \Phi_a\left({\textbf{x}}\right)=-\frac{1}{\epsilon_0}\int_{0}^{R} \frac{3a^2}{\left(r^2+a^2\right)^{\frac{5}{2}}} \left[\rho\left(\textbf{x}\right)+\frac{r^2}{6}\nabla^2\rho+...\right]r^2dr+O\left(a^2\right)$ It follows from $\nabla^2 \Phi_a\left({\textbf{x}}\right)=\frac{1}{4\pi\epsilon_0}\int\rho\left(\textbf{x}'\right)\left(\frac{3a^ 2}{\left(r^2+a^2\right)^{\frac{5}{2}}}\right)d^3x'$ by Taylor expansion of the charge density $\rho(x^{\prime})$ about the point $x$ and then doing angular integration . I've difficulty working out the details of this . $r=x-x^{\prime}$ and $\Phi_a$ is a function from $R^{3}$ to $R^{1}$ called the electric potential.