Three identical, parallel conducting plates A, B and C are placed as shown. Switches S₁ and S₂ are open, and can connect A and C respectively to earth when closed. +Q charge is given to B.

 

Whether either A or C is earthed (but not both together), a parallel plate capacitor is formed with B, with $\pm Q$ charges on the inner surfaces. Hence, charge of amount +Q flows to the earth. When both are earthed together, A and C effectively become connected. The plates now form two capacitors in parallel, with capacitances in the ratio 1:2, and hence share charge Q in the same ratio. 
${\mathrm{C}}_1=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{2\mathrm{d}},{\mathrm{C}}_2=\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}\frac{{\mathrm{C}}_1}{{\mathrm{C}}_2}=\frac{1}{2}$
Now we can write, $\frac{{\mathrm{q}}_1}{{\mathrm{q}}_2}=\frac{{\mathrm{C}}_1}{{\mathrm{C}}_2}=\frac{1}{2}$
Because potential difference of both the capacitors is same and  ${\mathrm{q}}_1+{\mathrm{q}}_2=\mathrm{Q}$ .
Solving these we get $q_1=\frac{Q}{3}; q_2=\frac{2Q}{3}$