Let $g(x)=\frac{1}{4 \pi |x|}$, $x$ is in $\mathbb{R}^3$, then $h(x)=\nabla^2 g =\delta(x)$ in distribution, i.e.$\int f(x)h(x)dx = f(0) $ for arbitrary $f(x)$ with compactly support.
I don't know how to show the integral of $h(x)$ over $\mathbb{R}^3$ is $1$, so can't get $h(x)=\delta(x)$ in distribution.
First of all if $x\not = 0$ then $h = \nabla^2 g = \nabla\cdot \left(\frac{x}{4\pi|x|^3}\right) = 0$. This means that
$$\int_{\mathbb{R}^3} \nabla^2 g d^3x = \int_{B(0,r)} \nabla^2 g d^3x $$
where $B(0,r)$ is any sphere of radius $r>0$ around the origin. Next use the Divergence theorem to find
$$\int_{B(0,r)} \nabla^2 g d^3x = \int_{B(0,r)} \nabla\cdot\left(\frac{x}{4\pi|x|^3}\right)d^3x = \int_{\partial B(0,r)} \left(\frac{x\cdot \hat{n}}{4\pi|x|^3}\right) dS = \frac{r}{4\pi r^3}\cdot 4\pi r^2 = 1$$
since the integrand is constant over $\partial B(0,r)$.