1. Given data

• Two wires of equal length $L$
• Diameter of each wire: $d = 2\ \text{mm}$
• Radius: $r = 1\ \text{mm} = 10^{-3}\ \text{m}$
• Cross-sectional area: $A = \pi r^2 = \pi (10^{-3})^2 = \pi \times 10^{-6}\ \text{m}^2$
• Equivalent resistance in parallel: $R_{eq} = 3\ \Omega$

Resistivities:

• Iron: $12\ \mu\Omega\text{ cm}$
• Copper-nickel alloy: $51\ \mu\Omega\text{ cm}$

Convert into SI units:

Since $1\ \mu\Omega\text{ cm} = 10^{-6}\Omega \times 10^{-2}\text{ m} = 10^{-8}\ \Omega\text{ m}$

Therefore, $\rho_1 = 12 \times 10^{-8} = 1.2 \times 10^{-7}\ \Omega\text{ m}$ $\rho_2 = 51 \times 10^{-8} = 5.1 \times 10^{-7}\ \Omega\text{ m}$

2. Resistance of each wire

Using $R = \frac{\rho L}{A}$

So, $R_1 = \frac{\rho_1 L}{A}, \qquad R_2 = \frac{\rho_2 L}{A}$

Since they are connected in parallel, $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$

Substitute: $\frac{1}{3} = \frac{A}{\rho_1 L} + \frac{A}{\rho_2 L}$

$\frac{1}{3} = \frac{A}{L}\left(\frac{1}{\rho_1} + \frac{1}{\rho_2}\right)$

Hence, $L = 3A\left(\frac{1}{\rho_1} + \frac{1}{\rho_2}\right)$

3. Substitute values

First compute: $\frac{1}{\rho_1} = \frac{1}{1.2\times 10^{-7}} = 8.333\times 10^6$ $\frac{1}{\rho_2} = \frac{1}{5.1\times 10^{-7}} = 1.961\times 10^6$

So, $\frac{1}{\rho_1}+\frac{1}{\rho_2} = 10.294\times 10^6$

Now, $L = 3 \times (\pi \times 10^{-6}) \times (10.294 \times 10^6)$

$L = 3\pi \times 10.294$

$L \approx 97.0\ \text{m}$

4. Match with options

The required equal length is $\boxed{97\ \text{m}}$

So the correct option is B.