A cube of metal is subjected to a hydrostatic pressure of 4 GPa.

(4) Bulk modulus, $B=\frac{P}{\frac{\Delta V}{V}}$

$\Rightarrow \frac{\Delta V}{V}=\frac{P}{B}$       ...(i)

If the side of cube is $L$ then $V=L^{3}$

$\frac{\Delta V}{V}=\frac{3 \Delta L}{L}=\frac{P}{B}$

$\Rightarrow \frac{\Delta L}{L}=\frac{1}{3} \times \frac{P}{B}=\frac{4 \times 10^{9} \mathrm{~Pa}}{3 \times 8 \times 10^{10} \mathrm{~Pa}}=\frac{1}{60}$

$\Rightarrow \frac{\Delta L}{L} \times 100=\frac{1}{60} \times 100=1.67 \%$