Drift speed of electrons, when 1.5 A current flows in a copper wire of cross section 5 mm$$^2$$ is $$v_d$$. If the electron density in copper is $$9 \times 10^{28}$$ m$$^{-3}$$, the value of $$v_d$$ in mm s$$^{-1}$$ is close to (Take charge of an electron to be $$= 1.6 \times 10^{-19}$$ C)

We have a current $$I = 1.5\ \text{A}$$ flowing through a copper wire of circular cross-sectional area $$A = 5\ \text{mm}^2$$. First we convert the area into SI units:

$$1\ \text{mm}^2 = 10^{-6}\ \text{m}^2 \; \Longrightarrow \; A = 5 \times 10^{-6}\ \text{m}^2.$$

The electron (number) density in copper is given as $$n = 9 \times 10^{28}\ \text{m}^{-3}.$$ The charge of a single electron is $$e = 1.6 \times 10^{-19}\ \text{C}.$$

The basic relation connecting the drift speed $$v_d$$ with these quantities is

$$I = n\,e\,A\,v_d.$$

We want $$v_d,$$ so we rearrange the formula:

$$v_d = \frac{I}{n\,e\,A}.$$

Substituting the numerical values step by step, we first evaluate the product $$n\,e$$:

$$n\,e = \left(9 \times 10^{28}\right)\left(1.6 \times 10^{-19}\right) = 14.4 \times 10^{9} = 1.44 \times 10^{10}\ \text{C m}^{-3}.$$

Now we include the area $$A = 5 \times 10^{-6}\ \text{m}^2$$:

$$n\,e\,A = \left(1.44 \times 10^{10}\right)\left(5 \times 10^{-6}\right) = 7.2 \times 10^{4}\ \text{C m}^{-2}.$$

Dividing the current by this product gives the drift speed:

$$v_d = \frac{I}{n\,e\,A} = \frac{1.5}{7.2 \times 10^{4}} = 2.0833 \times 10^{-5}\ \text{m s}^{-1}.$$

To express this speed in millimetres per second, we recall that $$1\ \text{m} = 1000\ \text{mm}$$:

$$v_d = 2.0833 \times 10^{-5}\ \text{m s}^{-1} \times 1000\ \frac{\text{mm}}{\text{m}} = 2.0833 \times 10^{-2}\ \text{mm s}^{-1}.$$

Numerically, $$2.0833 \times 10^{-2}\ \text{mm s}^{-1} \approx 0.02\ \text{mm s}^{-1}.$$

Hence, the correct answer is Option D.