= Solution
Write $C=A\sqrt{\pi/(2k)}$, so the radial field is $\Phi=C e^{-kr}/r$. Integrate its equation over a ball of radius $\epsilon$ centred at the source. The <divergence theorem> gives
$$
4\pi\epsilon^2\Phi'(\epsilon)-k^2\int_{r<\epsilon}\Phi\,dV=1.
$$
The second term is $4\pi k^2C\int_0^\epsilon re^{-kr}dr=O(\epsilon^2)$ and vanishes in the limit. Meanwhile
$$
4\pi\epsilon^2\Phi'(\epsilon)=-4\pi C(1+k\epsilon)e^{-k\epsilon}\longrightarrow-4\pi C.
$$
Thus
$$
\boxed{C=-\frac1{4\pi},\qquad A=-\frac1{4\pi}\sqrt{\frac{2k}{\pi}},\qquad\Phi(r)=-\frac{e^{-kr}}{4\pi r}.}
$$
The minus sign follows from the convention $(\Delta-k^2)\Phi=\delta^{(3)}$, opposite to the positive <static Yukawa Green function> for $-\Delta+k^2$. For arbitrary real $k$, replace it in the exponential by $|k|$ to select decay at infinity. The $k=0$ limit is $-1/(4\pi r)$.
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