Given k1 = k2 = 40 Nm⁻¹.

Springs $S_1$ and $S_2$ are connected in parallel. For springs in parallel, the effective spring constant is the sum of their individual spring constants.
Given $k_1 = k_2 = 40 \, \text{Nm}^{-1}$.

$$k_{12} = k_1 + k_2$$ $$k_{12} = 40 \, \text{Nm}^{-1} + 40 \, \text{Nm}^{-1}$$ $$k_{12} = 80 \, \text{Nm}^{-1}$$

The combination of $S_1$ and $S_2$ (with effective constant $k_{12}$) is connected in series with spring $S_3$. For springs in series, the reciprocal of the total effective spring constant is the sum of the reciprocals of the individual spring constants.
Given $k_3 = 40 \, \text{Nm}^{-1}$.

$$\frac{1}{k_{total}} = \frac{1}{k_{12}} + \frac{1}{k_3}$$ $$\frac{1}{k_{total}} = \frac{1}{80 \, \text{Nm}^{-1}} + \frac{1}{40 \, \text{Nm}^{-1}}$$

To add the fractions, find a common denominator:

$$\frac{1}{k_{total}} = \frac{1}{80 \, \text{Nm}^{-1}} + \frac{2}{80 \, \text{Nm}^{-1}}$$ $$\frac{1}{k_{total}} = \frac{3}{80 \, \text{Nm}^{-1}}$$ $$k_{total} = \frac{80}{3} \, \text{Nm}^{-1}$$

Using Hooke's Law, $F = kx$, where $F$ is the applied force (weight), $k$ is the total effective spring constant, and $x$ is the total extension.
Given $F = 45 \, \text{N}$.

$$x_{total} = \frac{F}{k_{total}}$$ $$x_{total} = \frac{45 \, \text{N}}{\frac{80}{3} \, \text{Nm}^{-1}}$$ $$x_{total} = 45 \times \frac{3}{80} \, \text{m}$$ $$x_{total} = \frac{135}{80} \, \text{m}$$

Simplify the fraction:

$$x_{total} = \frac{27}{16} \, \text{m}$$ $$x_{total} = 1.6875 \, \text{m}$$

The total extension of the system is $\text{1.6875 m}$.