In the steady state, all currents will have stopped, therefore we can ignore the resistors.
Note that both of the #3 mufarad# capacitors have a terminal that is common with the negative side of the battery (ignoring the #20 Omega# resistor). And the other terminal of those capacitors are both directly connected to point B. Therefore they are in parallel with each other yielding a capacitance of #6 mufarad#.
Note that both of the #1 mufarad# capacitors have a terminal that is common with the positive side of the battery (ignoring the #10 Omega# resistor). And the other terminal of those capacitors are both directly connected to point B. Therefore they are in parallel with each other yielding a capacitance of #2 mufarad#.
The charges are distributed so that the charges on the plates of the 2 pairs of capacitors directly connected to B are -q, +q, B, -q, +q left to right. The formula relating capacitance, charge, and voltage is
#V= q/C#.
The sum of the voltages #V_"AB" and V_"BC"# must be 100 V because the resistors have no current in them. The 2 equivalent capacitances are #6 mufarad and 2 mufarad# (left to right on the circuit diagram). The formula is #V= q/C#. Therefore the voltage #V_"BC"# must be the greater voltage by the ratio that 6 is greater than 2.
Therefore the answer is (3) #V_"AB" = 25 V and V_"BC" = 75 V#.
I hope this helps,
Steve
