A 27 mW laser beam has a cross-sectional area of 10 mm$$^2$$. The magnitude of the maximum electric field in this electromagnetic wave is given by: [Given permittivity of space $$\epsilon_0 = 9 \times 10^{-12}$$ SI units, Speed of light $$c = 3 \times 10^8$$ m/s]

We have a laser beam whose power (rate of energy transport) is given as $$P = 27 \text{ mW}$$. First, we convert this power into SI units:

$$P = 27 \text{ mW} = 27 \times 10^{-3}\, \text{W}$$

The cross-sectional area of the beam is given as $$10 \text{ mm}^2$$. Converting square millimetres to square metres:

$$1 \text{ mm} = 10^{-3} \text{ m} \quad \Longrightarrow \quad 1 \text{ mm}^2 = (10^{-3} \text{ m})^2 = 10^{-6} \text{ m}^2$$

Hence,

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

Intensity $$I$$ of an electromagnetic wave is the power per unit area, so we write

$$I = \dfrac{P}{A}$$

Substituting the numerical values,

$$I = \dfrac{27 \times 10^{-3}\, \text{W}}{1 \times 10^{-5}\, \text{m}^2}$$

$$I = 27 \times 10^{-3} \times 10^{5}\, \text{W m}^{-2}$$

$$I = 27 \times 10^{2}\, \text{W m}^{-2}$$

$$I = 2700\, \text{W m}^{-2}$$

Now, for a plane electromagnetic wave in free space, the relationship between the intensity and the maximum (peak) electric field $$E_0$$ is

$$I = \dfrac{1}{2}\, c\, \varepsilon_0\, E_0^{\,2}$$

where $$c$$ is the speed of light in vacuum and $$\varepsilon_0$$ is the permittivity of free space.

Rearranging for $$E_0$$, we get

$$E_0 = \sqrt{\dfrac{2I}{c\,\varepsilon_0}}$$

Substituting the numerical values $$I = 2700\,\text{W m}^{-2}$$, $$c = 3 \times 10^{8}\,\text{m s}^{-1}$$, and $$\varepsilon_0 = 9 \times 10^{-12}\,\text{F m}^{-1}$$, we have

$$E_0 = \sqrt{\dfrac{2 \times 2700}{(3 \times 10^{8})(9 \times 10^{-12})}}$$

First compute the numerator:

$$2 \times 2700 = 5400$$

Next compute the denominator $$c\,\varepsilon_0$$:

$$c\,\varepsilon_0 = (3 \times 10^{8})(9 \times 10^{-12})$$

$$= 27 \times 10^{-4}$$

$$= 2.7 \times 10^{-3}$$

So we now have

$$E_0 = \sqrt{\dfrac{5400}{2.7 \times 10^{-3}}}$$

Dividing inside the square root:

$$\dfrac{5400}{2.7 \times 10^{-3}} = \dfrac{5400}{2.7} \times 10^{3} = 2000 \times 10^{3} = 2.0 \times 10^{6}$$

Therefore,

$$E_0 = \sqrt{2.0 \times 10^{6}}$$

$$E_0 = \sqrt{2}\,\times 10^{3}\, \text{V m}^{-1}$$

$$E_0 \approx 1.414 \times 10^{3}\, \text{V m}^{-1}$$

$$E_0 \approx 1.4 \times 10^{3}\, \text{V m}^{-1}$$

Since $$1\,\text{kV m}^{-1} = 10^{3}\,\text{V m}^{-1}$$, we have

$$E_0 \approx 1.4\,\text{kV m}^{-1}$$

Hence, the correct answer is Option D.