Continue but use power 11 not 13 for the young modulus

The diameter of the wire is $d = 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m}$.
The radius is $r = d/2 = 1.0 \times 10^{-3} \text{ m}$.

$$A = \pi r^2$$ $$A = \pi (1.0 \times 10^{-3} \text{ m})^2$$ $$A = \pi (1.0 \times 10^{-6} \text{ m}^2)$$ $$A \approx 3.14159 \times 10^{-6} \text{ m}^2$$

Since the wires are joined end-to-end and the same load is applied, the stress in each wire is the same.
Stress ($\sigma$) = Force ($F$) / Area ($A$)

$$\sigma = \frac{200 \text{ N}}{3.14159 \times 10^{-6} \text{ m}^2}$$ $$\sigma \approx 6.366 \times 10^{7} \text{ Pa}$$

The stress in each wire is $\boxed{\text{6.37} \times 10^{7} \text{ Pa}}$.

2. Calculate the total extension of the composite wire.

Young's Modulus ($E$) = Stress ($\sigma$) / Strain ($\epsilon$)
Strain ($\epsilon$) = Extension ($\Delta L$) / Original Length ($L$)
So, $E = \frac{F/A}{\Delta L/L} = \frac{FL}{A \Delta L}$.
Rearranging for extension: $\Delta L = \frac{FL}{AE}$.

For copper wire:

$$\Delta L_c = \frac{F L_c}{A E_c}$$ $$\Delta L_c = \frac{(200 \text{ N})(1.5 \text{ m})}{(3.14159 \times 10^{-6} \text{ m}^2)(1.1 \times 10^{11} \text{ Pa})}$$ $$\Delta L_c = \frac{300 \text{ N}\cdot\text{m}}{3.45575 \times 10^{5} \text{ N}}$$ $$\Delta L_c \approx 8.681 \times 10^{-4} \text{ m}$$

For steel wire:

$$\Delta L_s = \frac{F L_s}{A E_s}$$ $$\Delta L_s = \frac{(200 \text{ N})(1.5 \text{ m})}{(3.14159 \times 10^{-6} \text{ m}^2)(2.0 \times 10^{11} \text{ Pa})}$$ $$\Delta L_s = \frac{300 \text{ N}\cdot\text{m}}{6.28318 \times 10^{5} \text{ N}}$$ $$\Delta L_s \approx 4.774 \times 10^{-4} \text{ m}$$

$$\Delta L_{\text{total}} = \Delta L_c + \Delta L_s$$ $$\Delta L_{\text{total}} = 8.681 \times 10^{-4} \text{ m} + 4.774 \times 10^{-4} \text{ m}$$ $$\Delta L_{\text{total}} = 13.455 \times 10^{-4} \text{ m}$$ $$\Delta L_{\text{total}} \approx 1.346 \times 10^{-3} \text{ m}$$

The total extension of the composite wire is $\boxed{\text{1.35} \times 10^{-3} \text{ m}}$.