= Solution
This part is a radial differential-equation identity. Expanding the spherical derivative gives
$$
\phi''+\frac2r\phi'+B^2\phi=0.
$$
Put $\psi=r\phi$. Since $\psi''=r\phi''+2\phi'$, multiplication of the equation by $r$ yields $\psi''+B^2\psi=0$. Its real solution is
$$
\psi=A\sin(Br)+C\cos(Br),\qquad
\boxed{\phi_{\rm cf}=A\frac{\sin(Br)}r+C\frac{\cos(Br)}r.}
$$
For a regular field at the centre, the cosine-over-radius contribution must vanish. The sine-over-radius term is regular because $\sin(Br)/r\to B$ as $r\to0$.
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