A highly resistive cube

If an electrical network is created using the complete graph $K_n$ , with $1 \,\Omega$ resistors on each edge, then the effective resistance $R_n$ between any two vertices is the same (by symmetry), and the effective graph resistance of $K_n$ is $n-1$ , so that ${n \choose 2}R_n \; = \; n-1$ which means that $R_n = 2n^{-1}$ . In this case ( $n=8$ , with $1$ k $\Omega$ resistors), we see that the effective resistance is $\tfrac28 \times 1000 = \boxed{250} \,\Omega$ .

If you apply $1$ V to one node and ground another node, then by symmetry, all the remaining nodes will be at $0.5$ V. So the total current going out through the grounded node will be $0.5$ times the number of nodes that are at $0.5$ V added to $1$ times the node to which $1$ V was applied. So $I = (6*0.5 + 1*1)/1000 = 1/250$ . So, the effective resistance will be: $R_e = \boxed{250} \mbox{ohms}$