Wires With Will

The total resistance was $48 \space \Omega$ . And it was divide into 48 equal parts. So, resistance of each piece will be $1 \space \Omega$ .

Node $A$ and node $Z$ (at the bottom, not shown), are two opposite nodes and this will be followed further in the solution.

Now, the current $I$ travels from node $A$ to node $Z$ (not shown in the figure). At node $A$ the current divide it selves into four parts into the branch ${(AC), (AE), (AG), (AI)}$

Since the resistances are equal in the above branches, the current also divides equally i.e. $\frac{I}{4}$ current from each branch. Now, these branches will be at same potential and because of that the branches in between the nodes $B, C, D, E, F, G, H, I$ becomes dead and current will not pass through them.

The current now passes through the branches $(CL), (EN)$ and other two branches(connected by nodes $G$ and $I$ ) not shown in the figure the current is still $\frac{I}{4}$ .

Again by method of equipotential surface, the branches in between nodes $K,L,M,N,O,P,Q,R$ are dead.

The solid is symmetrical about the set of points $K,L,M,N,O,P,Q,R$ , so same procedure has to be followed. And we will see the current never divides itself again and continues to flow as $\frac{I}{4}$ untill it reaches node $Z$ , where all the four branches carrying the current meet.

Now, while doing the traversing we encountered $16$ resistance arranged in group of four which are parallel to each other and those are connected by set of nodes $(A,C,L,V,Z),(A,E,N,X,Z)$ and two set of other five nodes not shown in the figure.

So, $R=\large \underbrace{\frac{4}{4}}_{4 \space parallel \space resistance}=1$

So, $100R=\color{#3D99F6}{\boxed{100}}$

By symmetry the above circuit can be simplified as:-

This can be simplified further using Wheatstone bridge.

The equivalent resistance comes out to be $\frac{4}{5}\Omega$ . So the answer is $\boxed {80}$