= Solution
Write $\mathbf r=R\boldsymbol\theta$, $|\boldsymbol\theta|=1$. Uniformly for $\mathbf r'$ in the source ball,
$$
|\mathbf r-\mathbf r'|=R-\boldsymbol\theta\cdot\mathbf r'+O(r_0^2/R),\qquad
|\mathbf r-\mathbf r'|^{-1}=R^{-1}+O(r_0/R^2).
$$
For fixed $k$ and $r_0$, the <outgoing Green function for the three-dimensional Helmholtz equation> therefore has the <far-field approximation for an outgoing source>
$$
G_k(R\boldsymbol\theta,\mathbf r')
=\frac{e^{ikR}}{4\pi R}e^{-ik\boldsymbol\theta\cdot\mathbf r'}+O(R^{-2}).
$$
Substitution into the source integral yields
$$
\psi(R\boldsymbol\theta,k)=\frac{e^{ikR}}R f_\infty(\boldsymbol\theta,k)+O(R^{-2}),\qquad
\boxed{f_\infty(\boldsymbol\theta,k)
=-\frac1{4\pi}\int_A Q(\mathbf r')e^{-ik\boldsymbol\theta\cdot\mathbf r'}\,d^3\mathbf r'.}
$$
Thus the <source-to-far-field operator at fixed frequency> is
$$
\boxed{T:L^2(A)\longrightarrow L^2(S_1),\qquad
(TQ)(\boldsymbol\theta)=-\frac1{4\pi}\int_AQ(\mathbf r')e^{-ik\boldsymbol\theta\cdot\mathbf r'}\,d^3\mathbf r'.}
$$
It samples the <Fourier transform> of the source on the sphere of radius $k$. Its square-integrable kernel makes it a <Hilbert-Schmidt operator>, hence a <compact operator>. In fact
$$
\|T\|_{\rm HS}^2=\frac{|A||S_1|}{16\pi^2}
=\frac{|A|}{4\pi}=\frac{r_0^3}{3}.
$$
Use complex <inner products> linear in the first argument: $(f,g)=\int f\overline g$. Interchanging the volume and surface integrals gives
$$
(TQ,g)_{L^2(S_1)}
=\int_A Q(\mathbf r')\overline{\left[
-\frac1{4\pi}\int_{S_1}g(\boldsymbol\theta)e^{ik\boldsymbol\theta\cdot\mathbf r'}\,dS(\boldsymbol\theta)\right]}\,d^3\mathbf r'.
$$
Hence \b[the <adjoint source-to-far-field operator> is]
$$
\boxed{(T^*g)(\mathbf r')=-\frac1{4\pi}\int_{S_1}
g(\boldsymbol\theta)e^{ik\boldsymbol\theta\cdot\mathbf r'}\,dS(\boldsymbol\theta).}
$$
This is a normalized <Herglotz wave function>; the change in exponential sign comes from the <complex conjugation> in the <adjoint operator>.
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