= Solution
A <fundamental solution of the Laplace equation> satisfies $\Delta_{\mathbf r}G_0=\delta(\mathbf r-\mathbf r')$ in the sense of <distributions>, without a boundary condition. Choose it rotationally symmetric about the source. Away from the source, with $s=|\mathbf r-\mathbf r'|$, the radial <Laplace equation> is $(sG_0')'=0$, giving $G_0=A\log s+B$.
Integrating over a small disc around the source and using the <divergence theorem> fixes the normalization:
$$
1=\int_{s<\varepsilon}\Delta G_0\,dS
=\int_{s=\varepsilon}\frac{\partial G_0}{\partial s}\,dC
=\frac A\varepsilon\,2\pi\varepsilon=2\pi A.
$$
Therefore
$$
\boxed{G_0(\mathbf r;\mathbf r')=\frac1{2\pi}\log|\mathbf r-\mathbf r'|+B.}
$$
The additive constant is arbitrary; adding other entire <harmonic functions> is also possible if rotational symmetry is not imposed. The positive sign here follows from the convention $\Delta G_0=+\delta$.
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