Given data:

Mass of the traffic light, $m = 45 \text{ kg}$

Radius of the steel wire, $r = 0.5 \text{ cm} = 0.005 \text{ m}$

Angle with the horizontal, $\theta = 15^\circ$

Young's modulus of steel, $Y = 2 \times 10^{11} \text{ Nm}^{-2}$

Acceleration due to gravity, $g = 9.8 \text{ ms}^{-2}$

Weight of the traffic light, $W = mg = 45 \times 9.8 = 441 \text{ N}$

Applying Lami's theorem at the junction where the wires meet:

$$ \begin{aligned} \frac{T_1}{\sin(90^\circ + 15^\circ)} &= \frac{W}{\sin(180^\circ - (90^\circ + 15^\circ) - (90^\circ + 15^\circ))} \\ \frac{T_1}{\cos 15^\circ} &= \frac{W}{\sin 150^\circ} \end{aligned} $$

Since the two wires are identical and symmetric, $T_1 = T_2 = T$.

$$ \begin{aligned} T &= \frac{W \cos 15^\circ}{\sin 150^\circ} \\ &= \frac{441 \times 0.9659}{0.5} \\ &= 852.36 \text{ N} \end{aligned} $$

The cross-sectional area of the wire is:

$$ \begin{aligned} A &= \pi r^2 \\ &= \pi \times (0.005)^2 \\ &= 7.854 \times 10^{-5} \text{ m}^2 \end{aligned} $$

The longitudinal strain is given by $\frac{\Delta L}{L} = \frac{T}{AY}$. Substituting the values:

$$ \begin{aligned} \frac{\Delta L}{L} &= \frac{852.36}{(7.854 \times 10^{-5}) \times (2 \times 10^{11})} \\ &= \frac{852.36}{1.5708 \times 10^7} \\ &= 5.426 \times 10^{-5} \end{aligned} $$

Thus, the strain in the wires is $5.426 \times 10^{-5}$.