The relationship between stress (σ), force (F), and area (A) is given by :

$$\sigma = \frac{F}{A}$$

In this context, the force is equal to the weight of the load, so we can substitute force with mass (m) times gravity (g) :

$$F = m \cdot g$$

From this, we get the formula for the cross-sectional area required to support a given mass :

$$A = \frac{F}{\sigma} = \frac{m \cdot g}{\sigma}$$

We can set up a proportionality relationship between the area for 10 metric tons (A₁₀) and the area for 25 metric tons (A₂₅) as follows :

$$\frac{A_{10}}{A_{25}} = \frac{m_{10}}{m_{25}}$$

Using the given values :

• $A_{10} = 2.5 \times 10^{-4}$ m²,
• $m_{10} = 10,000$ kg,
• $m_{25} = 25,000$ kg,

Solving for $A_{25}$ :

$$A_{25} = A_{10} \times \left(\frac{m_{25}}{m_{10}}\right) = 2.5 \times 10^{-4} \, \mathrm{m}^{2} \times \left(\frac{25,000 \, \mathrm{kg}}{10,000 \, \mathrm{kg}}\right) = 6.25 \times 10^{-4} \, \mathrm{m}^{2}$$

So, Option A $ (6.25 \times 10^{-4} \, \mathrm{m}^{2}) $ is the correct answer.