Three plates of common surface area A are connected as shown. The effective capacitance will be:

The given circuit is equivalent to parallel parallel combination of two identical capacitors, each having capacitance $\mathrm{C}=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}$.
Hence ${\mathrm{C}}_{\mathrm{eq}}=2\mathrm{C}=\frac{2{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}$