The work done in stretching a spring is equal to the area under the force-displacement graph. The total work done is the sum of the work done in the two segments of the graph, OA and AB.

1. Work done in the OA segment:

In the interval from $x=0$ to $x=0.3\text{ m}$, the force increases linearly from $0$ to $300\text{ N}$. The work done ($W_A$) is the area of the triangle formed by the origin and point A:

$$ \begin{aligned} W_A &= \frac{1}{2} \times \text{base} \times \text{height} \\ &= \frac{1}{2} \times 0.3\text{ m} \times 300\text{ N} \\ &= 45\text{ J} \end{aligned} $$

2. Work done in the AB segment:

In the interval from $x=0.3\text{ m}$ to $x=1.0\text{ m}$, the force increases from $300\text{ N}$ to $400\text{ N}$. The work done ($W_B$) is the area of the trapezoid formed under the line segment AB:

$$ \begin{aligned} W_B &= \frac{1}{2} \times (F_A + F_B) \times (x_B - x_A) \\ &= \frac{1}{2} \times (300\text{ N} + 400\text{ N}) \times (1.0\text{ m} - 0.3\text{ m}) \\ &= \frac{1}{2} \times 700\text{ N} \times 0.7\text{ m} \\ &= 350 \times 0.7 \\ &= 245\text{ J} \end{aligned} $$

Total work done:

$$ \begin{aligned} W &= W_A + W_B \\ &= 45\text{ J} + 245\text{ J} \\ &= 290\text{ J} \end{aligned} $$

The total work done in stretching the spring is $290\text{ J}$.