A cube of metal is subjected to a hydrostatic pressure 4 GPa. The percentage change in the length of the side of the cube is close to: (Given bulk modulus of metal, $$B = 8 \times 10^{10}$$ Pa)

We start with the definition of bulk modulus. For an isotropic body under a uniform (hydrostatic) pressure $$P$$, the bulk modulus $$B$$ is defined as

$$B \;=\; -\,\dfrac{P}{\dfrac{\Delta V}{V}}$$

Here $$\dfrac{\Delta V}{V}$$ is the volumetric (volume) strain and the minus sign accounts for the fact that an applied pressure (a compressive stress) makes the volume decrease (so $$\Delta V$$ is negative).

Rearranging the above formula to isolate the volume strain, we obtain

$$\dfrac{\Delta V}{V} \;=\; -\,\dfrac{P}{B}$$

Now we substitute the given numerical values. The hydrostatic pressure is $$P = 4 \text{ GPa} = 4 \times 10^{9}\ {\rm Pa}$$, and the bulk modulus is $$B = 8 \times 10^{10}\ {\rm Pa}$$. Therefore,

$$\dfrac{\Delta V}{V} \;=\; -\,\dfrac{4 \times 10^{9}}{8 \times 10^{10}}$$

Carrying out the division step by step, first divide the coefficients:

$$\dfrac{4}{8} = 0.5$$

Next handle the powers of ten:

$$\dfrac{10^{9}}{10^{10}} = 10^{-1} = 0.1$$

Multiplying the two intermediate results,

$$0.5 \times 0.1 = 0.05$$

Remembering the negative sign, we obtain

$$\dfrac{\Delta V}{V} = -\,0.05$$

This number means the volume decreases by 5 %. To convert this volume strain into linear (length) strain for a cube we use a geometrical relation. For a cube with side length $$l$$, the volume is $$V = l^{3}$$. If the side changes by a small amount $$\Delta l$$, we differentiate:

$$V = l^{3} \;\;\Longrightarrow\;\; \dfrac{dV}{V} = 3\,\dfrac{dl}{l}$$

In infinitesimal form, $$dV/V$$ is the volume strain and $$dl/l$$ is the linear strain. Re-writing this for finite but small changes, we approximate

$$\dfrac{\Delta V}{V} \approx 3\,\dfrac{\Delta l}{l}$$

Hence the linear strain is one-third of the volume strain:

$$\dfrac{\Delta l}{l} = \dfrac{1}{3}\,\dfrac{\Delta V}{V}$$

Substituting the value we already found,

$$\dfrac{\Delta l}{l} = \dfrac{1}{3}\,(-0.05)$$

Dividing by 3,

$$\dfrac{\Delta l}{l} = -0.016666\dots$$

To express this as a percentage change in length we multiply the absolute value by 100 %:

$$\bigl|\text{percentage change}\bigr| = 0.016666\dots \times 100\% = 1.6666\dots \%$$

Rounding to the number of significant figures implied by the data, we state

$$\text{percentage change in length} \approx 1.67\%$$

Hence, the correct answer is Option D.