Very good. Make sure that this does in fact satisfy the boundary conditions at all four electrodes and, of course, that it also satisfies Laplace's equation in the region between the electrodes.

Would you mind also helping with the next bit of the question

A solid, non-magnetic dielectric cylinder of radius R ≪ d and relative permittivity ǫr is placed between the electrodes, centred on the z axis. Sketch the distribution of surface polarization charge induced on the cylinder, and hence show that the cylinder does not acquire a dipole moment. By introducing a term of the form r−2 cos 2φ where appropriate (x = r cos φ, y = r sin φ), propose an approximate trial solution in cylin- drical coordinates for the potential inside and outside the cylinder, valid in the limit R/d → 0. Hence solve Laplace’s equation throughout the region between the electrodes in this limit.

For the first part I have
##\sigma=P \dot \hat{n}##
For a linear material
##P=\epsilon_0\chi_e E##
##E=-\nabla V = -\frac{2V_0}{d^2}x \hat{x} +\frac{2V_0}{d^2}y\hat{y} ##
##P=\frac{2\epsilon_0\chi_eV_0}{d^2}(x\hat{x}+y\hat{y})##
##\sigma = P \dot \hat{n} = \frac{2\epsilon_0\chi_eV_0}{d^2}(x\hat{x}+y\hat{y}) \dot (cos\theta \hat{x} + sin\theta \hat{y})##
using ##x=Rcos\theta## and ##y=Rsin\theta##
##\sigma = \frac{2\epsilon_0\chi_eV_0R}{d^2}##

Which is uniform all the way around so no dipole moment

The first part of the problem asks for a (rough?) sketch of the surface charge distribution. Then, use this sketch to deduce zero net dipole moment of this charge distribution. (That's my interpretation of what they want, anyway.)

Note that a uniform, nonzero ##\sigma## would imply a nonzero net charge on the cylinder.

Sketch E-field lines in the region of the origin. Look for symmetry. This should help you see any symmetry of the surface charge density.

Usually, it is assumed that the susceptibility ##\chi_e## is constant inside the material (i.e., does not depend on position). From ##\bf D = \epsilon_0 \left( 1 + \chi_e \right) \bf E##, deduce that ##\nabla \cdot \bf D## ## = \epsilon_0 \left( 1 + \chi_e \right)\nabla \cdot \bf E##. But we know ##\nabla \cdot \bf D## ## = 0## (assuming no free volume charge density ##\rho_f##). Hence, ##\nabla\cdot \bf E## ## = 0##. And this implies zero total volume charge density ##\rho_f + \rho_b##. Hence, what can you conclude about any bound charge density ##\rho_b##?