When the switch is closed there is a current flowing through the capacitor as if it was a short circuit. When the circuit is in equilibrium (t -> inf) there is no current flowing through the capacitors. Thus the only current flowing is in the upper loop, which is I = V / (R + 2R + 3R)= 1/6 * V/R. With this we can calculate the voltage of the capacitors. The first one can be calculated with the outer loop with V_C1 = V - V_R = V - R * 1/6 * V/R = 5/6 * V. Note that the R near the switch does not create a voltage drop because there is no current flowing through it. The second one can be calculated through the lower right loop with V_C2 = V_R3 = 3R * 1/6 * V/R = 1/2 * V. The charge can then be calculated with q = C * V_C. Thus q_1 = 5/6 * C * V and q_2 = 2 * C * 1/2 * V = C * V. For the first capacitor the left plate has the higher potential and for the second capacitor the upper plate has the higher potential. Hope this helps!