1. Use the resistance formula

For a wire, $R = \rho \frac{L}{A}$ where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

Since all wires are copper and of the same length, resistance is inversely proportional to area: $R \propto \frac{1}{A} \propto \frac{1}{r^2}$

2. Original wire

The original single wire has radius $r_1 = 9\text{ mm}$ and resistance $R_1 = 14\,\Omega$

3. Resistance of one new wire

Each new wire has radius $r_2 = 3\text{ mm}$

So, $\frac{R_2}{R_1} = \frac{r_1^2}{r_2^2} = \frac{9^2}{3^2} = \frac{81}{9} = 9$

Thus resistance of one thin wire is $R_2 = 9R_1 = 9 \times 14 = 126\,\Omega$

4. Seven identical wires in parallel

If 7 identical resistors of resistance $126\,\Omega$ are connected in parallel, equivalent resistance is $R_{\text{eq}} = \frac{126}{7} = 18\,\Omega$

5. Final answer

$\boxed{18\,\Omega}$

So the correct option is B.

6. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.