Given data

• Edge of copper cube: $a = 10\text{ cm} = 0.1\text{ m}$
• Pressure applied: $P = 7 \times 10^6\text{ Pa}$
• Bulk modulus of copper: $K = 1.4 \times 10^{11}\text{ Pa}$

Volume of the cube $V = a^3 = (0.1)^3 = 10^{-3}\text{ m}^3$

Use bulk modulus relation Bulk modulus is given by $K = -\frac{\Delta P}{\Delta V/V}$ Taking magnitude for volume contraction, $\frac{\Delta V}{V} = \frac{P}{K}$

Therefore, $\Delta V = V\frac{P}{K}$

Substitute values $\Delta V = 10^{-3} \times \frac{7\times 10^6}{1.4\times 10^{11}}$

$\frac{7}{1.4} = 5$ and $\frac{10^6}{10^{11}} = 10^{-5}$

So, $\Delta V = 10^{-3} \times 5\times 10^{-5} = 5\times 10^{-8}\text{ m}^3$

Convert to mm$^3$ Since $1\text{ m}^3 = 10^9\text{ mm}^3$ we get $\Delta V = 5\times 10^{-8} \times 10^9 = 50\text{ mm}^3$

Final answer $\boxed{50}$ mm$^3$