Since the sphere is perfectly diamagnetic, we have the condition that $\vec{B} = 0$ inside the sphere. By the principle of superposition, this means that the sphere has a magnetization $\vec{M}$ that induces a magnetic field $-\vec{B}_0$ inside the sphere. It turns out, this is satisfied for a uniform $\vec{M}$ pointing in $-\hat{B}_0$ direction, if you find the right magnitude for $\vec{M}$. To do so, use bound currents with the Biot-Savart Law to solve for the magnitude $M$. You'll find that $\vec{M}=-\frac{3}{2\mu_0}\vec{B}_0$. Finally, you can find the magnetic moment $\vec{m}$ by integrating this magnetization over the sphere's volume, giving you $\vec{m} = \frac{4}{3}\pi R^3 \vec{M} = -\frac{2\pi}{\mu_0} R^3 \vec{B}_0$.
Try doing the problem without the dielectric filling the sphere first as a warmup. Here's why that's useful:
Note that the definition of the displacement field is:$$\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P},$$ where $\mathbf{P}$ is the dipole moment density. Taking the curl of both sides of the equation in the static state gives:
$$\nabla \times \mathbf{D} = \nabla \times \mathbf{P}.$$ Taking the divergence gives:
$$\nabla \cdot \mathbf{D} = \rho + \nabla \cdot \mathbf{P}.$$ We usually define the bound charge density as $\rho_{\mathrm{bound}} \equiv -\nabla \cdot \mathbf{P}$, and the remaining charge density is then called 'free', $\rho_{\mathrm{free}}\equiv \nabla \cdot \mathbf{D}$.
The divergence and curl equations above can be converted into the boundary conditions for the three vector fields across surfaces. Applying Gauss's law to the divergence equations gives:
$$\begin{align}\Delta \mathbf{P} \cdot \hat{n} & = \sigma_{\mathrm{bound}}, \\
\Delta \mathbf{D} \cdot \hat{n} & = \sigma_{\mathrm{free}},\ \mathrm{and} \\
\Delta \mathbf{E} \cdot \hat{n} & = \frac{\sigma_{\mathrm{bound}}}{\epsilon_0}, \end{align}$$ where $\hat{n}$ is a unit vector perpendicular to the surface and the $\Delta$s are understood to be a difference on one side of the surface from the other. These conditions are frequently expressed in terms of the components of the field perpendicular to the surface (eg $\mathbf{E}_\perp$), as implied by the dot product. In words, these mean that the change perpendicular components of the fields are dictated by their associated surface charge density.
Unlike the electric field, though, the displacement and polarization can have solenoidal components, as you can see from the curl equation above. Applying Stoke's theorem at the boundary surface for them gives:
$$\begin{align}\Delta \mathbf{D}\times \hat{n} &= \Delta \mathbf{P}\times \hat{n}, \ \mathrm{and} \\
\Delta \mathbf{E} \times \hat{n} &= 0.\end{align}$$
Like the conditions on the perpendicular components, this can be expressed as a condition on the components of the fields parallel to the surface (eg $\mathbf{E}_{||}$). In words, the parallel components of the electric field must be continuous, and the discontinuity in the parallel components of the polarization must be balanced by a discontinuity in the parallel components of the displacement field.
Note that I don't think you actually need to find $\mathbf{D}$ to solve this problem - it should be sufficient to solve the dielectricless problem, use that to infer the polarization in the sphere, and from that calculate how the net surface charge density is modified by the bound charge.
All of that said, you should find that the empty sphere is a pure dipole field on the inside and the outside. Keep in mind that a dipole field appropriate for $r\rightarrow 0$, ie 'inside', has a different structure than one for $r\rightarrow \infty$, 'outside'. Your textbook should have a section on solving such problems using Legendre polynomials, as mentioned in a comment by @PrasadMani. The multipole fields are separated by the fact that they behave under rotations in orthogonal fashions, and it shouldn't surprise you to find that adding a rotationally invariant dielectric sphere won't cause a mixing of multipole field types.
Best Answer
The boundary conditions you should use are that as $r\rightarrow \infty$, $\mathbf{B}\rightarrow B_0\hat{\mathbf z}$; and that at $r=a$, $\mathbf B \cdot \hat{\mathbf r}=0$.