Capacitance

1.77nF

@100miles
Suppose the negative terminal of the battery gives a charge  Q to the plate B. As the situation is symmetric on the two sides of B, the two faces of the plates B will share equal chargesQ/2
each. As the positive terminal of the battery has supplied just this much charge (+ Q) to A and C, the outer surfaces of A and C will have no charge. The distribution will be as shown in figure. The capacitance between the plates A and B is
8.85 X 10^(12) Fm(1) X (200 X 10^(4) m^2)/(2X10^(4) m) = 8.85 X 10 ^(10) F = 0.885 nF. Thus, (Q)/(2) = 0.885 mF X 20 V = 17.7 nC. The distribution of charge on various surfaces may be the equivalent capacitance is
(Q)/(20 V) = 1.77 nF.