= Solution
The three-dimensional <fundamental solution of the Laplace equation>, with $\Delta F=+\delta$, is
$$
\boxed{F(\mathbf r,\mathbf r')=-\frac1{4\pi|\mathbf r-\mathbf r'|}.}
$$
It is <harmonic> away from the source, and its gradient has total outward <flux> one through a small source-centred sphere.
For $s=|\mathbf r'|\in(0,1)$, place the image at $\mathbf r^*=\mathbf r'/s^2$, outside the unit ball. On $|\mathbf r|=1$, expansion of squared distances gives $|\mathbf r-\mathbf r^*|=|\mathbf r-\mathbf r'|/s$. The <method of images> therefore gives the <Dirichlet Green function inside a sphere>
$$
\boxed{G_0(\mathbf r,\mathbf r')=-\frac1{4\pi}\left[\frac1{|\mathbf r-\mathbf r'|}-\frac1{s|\mathbf r-\mathbf r^*|}\right].}
$$
The boundary value is zero by the distance identity, and the image term is harmonic throughout the physical ball. Thus $\Delta G_0=\delta(\mathbf r-\mathbf r')$ there. At the central source, take the finite limit of the weighted image term:
$$
\boxed{G_0(\mathbf r,0)=-\frac1{4\pi|\mathbf r|}+\frac1{4\pi}.}
$$
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