1. Use bulk modulus relation

For hydrostatic pressure,

$B = -\frac{\Delta P}{\Delta V/V}$

So the fractional change in volume is

$\frac{\Delta V}{V} = -\frac{\Delta P}{B}$

Given:

$\Delta P = 4\,\text{GPa} = 4\times 10^9\,\text{Pa}$ $B = 8\times 10^{10}\,\text{Pa}$

Therefore,

$\left|\frac{\Delta V}{V}\right| = \frac{4\times 10^9}{8\times 10^{10}} = \frac{1}{20} = 0.05$

So the percentage decrease in volume is

$0.05\times 100 = 5\%$

2. Relate volume change to change in side length

Let the side of the cube be $a$. Then

$V = a^3$

For small changes,

$\frac{\Delta V}{V} = 3\frac{\Delta a}{a}$

Hence,

$\frac{\Delta a}{a} = \frac{1}{3}\frac{\Delta V}{V}$

Taking magnitude,

$\left|\frac{\Delta a}{a}\right| = \frac{0.05}{3} = 0.0167$

So the percentage change in side length is

$0.0167\times 100 = 1.67\%$

3. Match with options

The closest option is:

$\boxed{1.67}$

So, Option C is correct.