1. Energy density in elastic region

For a material obeying Hooke’s law, the elastic energy stored per unit volume is the area under the stress–strain graph:

$u = \int_0^{\varepsilon} \sigma \, d\varepsilon$

Since the graph is linear,

$u = \frac{1}{2}\sigma\varepsilon$

2. From the graph

At strain

$\varepsilon = 5\times 10^{-4}$

the corresponding stress from the given straight-line graph is

$\sigma = 100\times 10^6\ \text{Pa}$

3. Compute energy density

$u = \frac{1}{2}(100\times 10^6)(5\times 10^{-4})$

$u = \frac{1}{2}(5\times 10^4)$

$u = 2.5\times 10^4\ \text{J/m}^3$

4. Convert to kJ/m$^3$

$2.5\times 10^4\ \text{J/m}^3 = 25\ \text{kJ/m}^3$

Final Answer

$\boxed{25}$

This matches the stored correct answer.