Let us be more specific about electro-thermal analogy: temperature T corresponds to electrostatic potential \varphi and heat flux density \boldsymbol J corresponds to the electric D-field \boldsymbol D (bold symbols denote vectors). Indeed, if we compare equations which define the solutions, \boldsymbol D =-\varepsilon \textbf{{grad}} \varphi with \oint \boldsymbol D \mathrm{d} \boldsymbol A =q, and \boldsymbol J=-\kappa \textbf{{grad}} T with \oint \boldsymbol J \mathrm{d} \boldsymbol A =P, respectively (with q denoting the full unbound charge inside the integration surface, and P – the total heat power released from therein), we’ll see that there is a perfect match, with \varepsilon\equiv \varepsilon (\boldsymbol r) corresponding to \kappa\equiv\kappa (\boldsymbol r).
So, first you need to find the field induced around a dielectric ball which is surrounded by an homogeneous D-field (that field appears to be a perfect dipole field); the total field outside the sphere will be the superposition of the homogeneous field and the dipole field.
Next you need to be very careful with the boundary conditions: constant temperature at the plates means constant voltage between the plates. However, when the magnitude of the dipole moment on the sphere is changing, the changing dipole field will induce a certain voltage change across the plates! So, your next task is to figure out by how much is the voltage changed. This is not exactly a trivial task: while you can make use of the Hint 2, there is still one more step to be done to reach the answer (more details about that in the upcoming Hint 4).