A

Step 1: Identify the resistors and their configuration from point A to B. The circuit is composed of a combination of series and parallel resistances.
Step 2: Start by simplifying the resistors in series and parallel step by step. The 10Ω and 10Ω resistors are in series which gives an equivalent resistance of 20Ω.
Step 3: Next, we have a parallel combination of 20Ω (from the previous step) and 3Ω. The equivalent resistance for parallel resistors is given by:
\[ R_{eq1} = \frac{1}{{\frac{1}{20} + \frac{1}{3}}} = \frac{60}{11} \approx 5.45Ω. \]
Step 4: Now, adding the next resistors in series again: 5Ω + 8Ω = 13Ω.
Step 5: There are now two parallel branches: one branch is 5.45Ω and the other is 13Ω. Computing the equivalent of these two:
\[ R_{eq2} = \frac{1}{{\frac{1}{5.45} + \frac{1}{13}}} \approx 4.14Ω. \]
Step 6: Finally, we add the 6Ω resistor in series to this equivalent resistance:
\[ R_{eq} = 4.14 + 6 = 10.14Ω. \] Therefore, the final equivalent resistance between A and B is approximately 10.14Ω.