A physically unrealizable problem

Considering the above figure as a cube having each resistor of $R\Omega$ and a voltage source of $EMF (V)$ connected across $A$ and $B$ .

$\color{#3D99F6}{V}=IR_{AB}$

By applying Kirchhoff's laws on loop $ACDBA$ ,

$\color{#3D99F6}{V}=\frac{I}{3}R+\frac{I}{6}R+\frac{I}{3}R \\ IR_{AB}=IR\left(\frac{2}{3}+\frac{1}{6}\right) \\ R_{AB}=\frac{5}{6}R$

So, we can get the $R_{eq}$ for 1-order hexaresistor as $\left(\frac{5}{6}\right)1000 \Omega$ .

And 2-order hexareaistor will have $R_{eq}$ as $\left(\frac{5}{6}\right)^2 1000 \Omega$ .

And so on.

So our answer is as follows,

$=\left(\frac{5}{6}\right)1000+ \left(\frac{5}{6}\right)^2 1000+ \left(\frac{5}{6}\right)^3 1000+\ldots \\ =1000\left(\frac{\frac{5}{6}}{1-\frac{5}{6}} \right) \\ =1000×5 = \boxed{5000}$

A beauty of coincidence!

Each R at the 4 corners are converted from R into 2 R // 2 R as the very first start!

We obtain a proposed equivalent circuit of 4 main parallels yet to be verified as follows:

|---- 2 R -- $*$ - R ---- $*$ --- 2 R ----| [1] $$ |~~~~~~---- R - $*$ -- 2 R ----|

|---- R ---- |~~~~~~~~~| [2]

|~~~~~~---- R -#-- 2 R ----| $$ |---- 2 R -- $*$ - R ----~~~~~~|

|~~~~~~~~~|---- R ----| [3]

|---- 2 R --#- R ----~~~~~~| $$ |---- 2 R ---#---- R -#-- 2 R ----| [4]

Note that "~" is a sign I applied as a push, it stands for a space. 2 detached points $*,$ and 2 detached points # changed into 4 $\times *$ and 4 $\times$ #respectively.

With test voltage of 5 V on rail on our left, and 0 V on rail on our right, let's see partial circuits [1] and [2]:

$[1]$ : 5 V $\rightarrow$ 3 V $\rightarrow$ 2 V $\rightarrow$ 0 V as $\displaystyle \frac{2 R}{2 R + R +2 R} 5 V = 2 V$

$[2]$ : 5 V $\rightarrow$ 3 V $\rightarrow$ 2 V $\rightarrow$ 0 V as $\displaystyle \frac{1.5 R}{R + 1.5 R} 5 V = 3 V$ and $\displaystyle \frac{2 R}{R + 2 R} 3 V = 2 V$

The coincidence here is 2 V onto detached points between [1] and [2] and also same detached points between [3] and [4] are allowed as there shall be no current to flow between them no matter there are contacts or not, because of same voltage or zero potential difference!

Consequently, we have obtained 5 R // 2.5 R // 2.5 R // 5 R which is $\Large \frac{1}{\frac15 + \frac{1}{2.5} + \frac{1}{2.5} + \frac15}$ $R$ = $\Large \frac56$ $R$

R<1> = 1000 $\Omega \times \frac56$ = $833 \frac13 \Omega$

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833.3333333
694.4444444
578.7037037
482.2530864
401.877572
334.8979767
279.0816472
232.5680394
193.8066995
161.5055829
...
9.08359E-21
7.56966E-21
6.30805E-21
5.25671E-21
4.38059E-21
3.65049E-21
3.04208E-21
2.53506E-21
2.11255E-21
1.76046E-21

The above is R<1>, R<2>, R<3>, R<4>, ..., R<300>; before R< $\infty$ >, it has approached to 0 $\Omega!$

Each of which multiplied by $\frac56$ for each step of resistance followed.

Sum of R<i> from 1 to $\infty$ = 1000 $\Large \frac{\frac56}{1 - \frac56}$ $\Omega$ = 5000 $\Omega$

Answer: $\boxed{5000}$

what an amazing question..... really took me some time to figure out :)