C

Step 1: Understand the Configuration
The given figure represents an infinite number of equilateral triangles, with each subsequent triangle having sides that are halved from the previous one. The original side length AB is given as 'a'.

Step 2: Find the Resistance of the Wire
The resistance of a wire segment is given by the formula:
$$ R = \rho \frac{L}{A} $$
where:
- $\rho$ is the resistivity
- $L$ is the length of the wire
- $A$ is the cross-sectional area (constant for this problem).

Since the triangles are equilateral, each triangular wire segment's length can be calculated based on its side length.

Step 3: Calculate Total Resistance
1. The resistance between points A and B for the first triangle (side length = a) can be expressed as $R_1 = \rho \frac{a}{A}$.
2. For the next triangle (side length = a/2), the resistance becomes $R_2 = \rho \frac{(a/2)}{A}$.
3. This pattern continues such that the resistance of the nth triangle can be represented as:
$$ R_n = \rho \frac{(a/2^{n-1})}{A} $$
4. The total resistance of an infinite series can be expressed as:
$$ R_{total} = R_1 + R_2 + R_3 + ... = \rho \frac{a}{A} \left(1 + \frac{1}{2} + \frac{1}{4} + ...\right) $$
5. The series $1 + \frac{1}{2} + \frac{1}{4} + ...$ converges to 2.

Step 4: Final Formula
Combining these results, we find:
$$ R_{AB} = \rho \frac{a}{A} \cdot 2 = \frac{2\rho a}{A} $$
where A is constant.

Thus, the final expression for the resistance between AB, with the given assumptions, converges to a function of the side 'a' and the resistivity 'ρ'. Therefore, based on common configurations and deducing towards standard results found through similar construction:

Therefore, the correct option is C.