Resistors!

First let us consider symmetry along $AB$

Since the arrangement is symmetric about $AB$ , we can fold the plane to make overlapping resistors connected parallely.

Now the folded image is symmetric about $CD$

So, $I_1 = I_2$ and $I_3 = I_4$ , and hence we can split the arrangement at $O$ .

Now, the arrangement can be further modified into:

So, $\begin{aligned}R_0 &= \dfrac R2 + \dfrac 1{\dfrac 2{3R} + \dfrac2{4R}} + \dfrac R2\\ &=\dfrac{6R}7 + R \\&= \dfrac{13}7R\end{aligned}$

$\Huge \therefore \boxed{\dfrac{R_0}R = \dfrac{13}7}$

$\color{#3D99F6}{\text{I have a slightly different solution.}}$

Due to symmetry, we note that points C, D, E, ... have the same potential with the respective C', D', E'. Therefore, they can be considered as same points and hence the equivalent circuit of $12 \times R||R$ or $\dfrac{R}{2}$ resistors.

Again due to symmetry we note that both the potentials of C and G are $\frac{1}{2}V_{AB}$ , therefore they can be considered as a same point and hence the second equivalent circuit with $2 \times \dfrac{R}{4}$ .

Therefore,

$\begin{aligned} R_0 & = 2 \times \left(\frac{R}{2} + \left(\frac{R}{2} + \frac{R}{4} \right) || \left(\frac{R}{2} + \frac{R}{2} \right) \right) = 2 \times \left(\frac{R}{2} + \left(\frac{3R}{4} \right) || \left( 1 \right) \right) \\ & = 2 \times \left(\frac{R}{2} + \frac{\frac{3R}{4} \times R}{\frac{3R}{4} + R} \right) = 2 \times \left(\frac{R}{2} + \frac{3R}{7} \right) = \frac{13}{7}R \\ \Rightarrow \frac{R_0}{R} & = \frac{13}{7} \end{aligned}$

$\Rightarrow a + b + ab = 13 + 7 + 13 \times 7 = \boxed{111}$

Thing that should have been clearer in description was $R$ as resistance onto side of smallest unit of square! Let this be a summary:

$R_0 =~$ [~ $R$ ~(3 $R$ // 4 $R$ )~ $R$ ~] // [~ $R$ ~(4 $R$ // 3 $R$ )~ $R$ ~] = [ $\frac{3 \times 4}{3 + 4} R + 2 R$ ] // [ $\frac{4 \times 3}{4 + 3} R + 2 R$ ] = $\frac{26}{7} R \times \frac12 = \frac{13}{7} R$

13 + 7 + 13 $\times$ 7 = 111

Answer: $\boxed{111}$