$$\begin{gathered} Force = Electric fieldx charge \\ Electric field in a capacitor = \frac{\sigma }{2{\epsilon }_o}, \sigma =\frac{Q}{A} \\ \therefore Force = \frac{Q}{2A{\epsilon }_o}xQ=\frac{Q^2}{2A{\epsilon }_o} \end{gathered}$$

Alternate solution:

The capacitance of the capacitor is $C={\epsilon }_{0}A/x$ where x is the distance between the plates. The energy stored in the capacitor is.

$U=\frac{1}{2}C{V}^2=\frac{{\epsilon }_{0}A{V}^2}{2x}$

Differentiating .w.r. t x we get 

$\frac{dU}{dx}=\frac{{\epsilon }_{0}A{V}^2}{2}\frac{d}{dx}\left(\frac{1}{x}\right)=-\frac{{\epsilon }_{0}A{V}^2}{2x^2}$

The force of attraction between the plates is 

$F=-\frac{dU}{dx}=\frac{{\epsilon }_{0}A{V}^2}{2x^2}$              (i)

Now $Q=CV=\frac{{\epsilon }_{0}AV}{x}$  or         (ii)

Using (ii) in (i) we get 

$F=\frac{Q^2}{2{\epsilon }_{0}A}$

Hence the correct choice is (c).