= Solution
Since there are no currents, $\nabla\times B=0$ in each constant-permeability region, while $\nabla\cdot B=0$ gives $\nabla^2\psi=0$. The displayed dipole forms are therefore harmonic. Regularity at the origin and the field at infinity give
$$
b_1=0,\qquad a_3=B_0.
$$
For $\psi=(ar+b/r^2)\cos\theta$,
$$
B_r=(a-2b/r^3)\cos\theta,
\qquad
B_\theta=-(a+b/r^3)\sin\theta.
$$
The <magnetic spherical-shell matching> conditions, continuity of $B_r$ and $H_\theta$, give the remaining four equations
$$
a_1-\frac{2b_1}{R_1^3}=a_2-\frac{2b_2}{R_1^3},
\qquad
\frac{a_1+b_1/R_1^3}{\mu_0}
=\frac{a_2+b_2/R_1^3}{\mu},
$$
$$
a_2-\frac{2b_2}{R_2^3}=a_3-\frac{2b_3}{R_2^3},
\qquad
\frac{a_2+b_2/R_2^3}{\mu}
=\frac{a_3+b_3/R_2^3}{\mu_0}.
$$
The two $R_1$ equations yield
$$
a_2=\frac{\mu_0+2\mu}{3\mu_0}a_1,
\qquad
b_2=\frac{\mu-\mu_0}{3\mu_0}R_1^3a_1.
$$
Together with $b_1=0$ and $a_3=B_0$, these are the requested expressions. Substituting them into the two $R_2$ equations gives a two-by-two linear system for $a_1$ and $b_3$.
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