How to solve the exterior problem on a ball with radius $r$ in the 3d space? I have to found u such that:
$\Delta u = 0$ in $B(0,r)^C$
The solution to the Dirichlet problem uses the Green's function for the Laplacian. In 3D, this is $H(r) = -1/4\pi|r|$ (for vector $r$), such that $\nabla^2 H(r) = \delta(r)$, per the definition of the Green's function.
Some use of the various generalized Stokes theorems gives the following (when $\nabla^2 u = 0$):
$$u(r) = \oint \nabla'H(r-r') \cdot dS' \; u(r') - \oint H(r-r') \; dS' \cdot \nabla' u(r')$$
The choice of the Dirichlet Green's function enforces that $H(r-r') = 0$ on the boundary. The second integral vanishes, leaving only the first. You'll have to be careful about the orientation of $dS'$ (since this is an exterior problem), but otherwise, you should be able to exploit the symmetries of the problem to reduce the surface integral to something tractable. Good luck!