Effective resistance from a random resistance network
A generalised picture of the resistor network (As I have understood it) is shown above. V1-V2 is the applied potential difference between the two faces (seems to be 20 V in your picture). V3, V4, V5, V6 etc are the node voltages you have already solved for.
To find the equivalent resistance, you need to divide the applied potential difference by the current flowing into the material. For that you need to sum up the currents of all the resistors which are directly connected to a selected face (say, the face with potential V1). R_eqv = V1 / (i1+i2+i3+i4).
To find those currents, divide the potential difference across each resistor (directly connected to the selected face) by its resistance. e.g. (V1-V3)/R1.
I don't think you need to worry about all the internal node voltages or any other internal details to calculate resistance since resistance is something which is seen by the measuring equipment which can only see the faces and the net current flowing into the  material.
I don't know if the above assumption would hold good if anything inside the material acts as a voltage source (With quantum and nano things, you can never tell :| ). If there is some such phenomenon, you can still find equivalent resistance by varying the applied voltage and dividing the difference in applied voltage by difference in current. e.g. find current for 20 V applied voltage. Then repeat for 21 V applied voltage. The equivalent resistance is then 1 V / (difference in current).