OK, your formula for ##\langle\phi(x,y)\rangle_c## is wrong*. I'll show you how it's derived. The formula for ##\langle\phi(x,y)\rangle_d## can be found in a similar way. I don't know how to show what you've been asked to show, but seeing where the other formulas come from may give you an idea of how to proceed.
The idea is to use a Taylor series to approximate ##\phi##. For example,
\begin{align*}
\phi(x+h,y) &= \phi(x,y) + h\frac{\partial\phi}{\partial x}(x,y) + \frac{1}{2!}h^2\frac{\partial^2\phi}{\partial x^2}(x,y) + \text{ higher-order terms} \\
\phi(x-h,y) &= \phi(x,y) - h\frac{\partial\phi}{\partial x}(x,y) + \frac{1}{2!}h^2\frac{\partial^2\phi}{\partial x^2}(x,y) + \text{ higher-order terms}.
\end{align*} If you add these two equations together, the first-order terms cancel, and you're left with
$$\phi(x+h,y) + \phi(x-h,y) = 2\phi(x,y) + h^2\frac{\partial^2\phi}{\partial x^2}(x,y) + \text{ higher-order terms}.$$ If you do the same thing with y, you end up with
$$\phi(x,y+h) + \phi(x,y-h) = 2\phi(x,y) + h^2\frac{\partial^2\phi}{\partial y^2}(x,y) + \text{ higher-order terms}.$$ Combining the x and y equations and neglecting the higher-order terms, you end up with
$$\phi(x+h,y) + \phi(x-h,y) + \phi(x,y+h) + \phi(x,y-h) = 4\phi(x,y) + h^2\left(\frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2}\right).$$ The term in the parentheses vanishes because ##\nabla^2\phi = 0##, so solving for ##\phi(x,y)##, you get
$$\phi(x,y) = \frac{1}{4}[\phi(x+h,y) + \phi(x-h,y) + \phi(x,y+h) + \phi(x,y-h)].$$ You can derive the formula for ##\langle\phi(x,y)\rangle_d## similarly by considering ##\phi(x\pm h, y\pm h)##.*It might be that your formula for ##\langle\phi(x,y)\rangle_c## is correct, and the missing term gives rise to the expression you're asked to prove. That's another idea you might want to investigate. I think, however, that you simply copied the formula down incorrectly in your notes.
