If the magnetic field of a plane electromagnetic wave is given by (The speed of light $$= 3 \times 10^8$$ m/s) $$B = 100 \times 10^{-6} \sin\left[2\pi \times 2 \times 10^{15}\left(t - \frac{x}{c}\right)\right]$$ then the maximum electric field associated with it is:

The magnetic field of the plane electromagnetic wave is written as

$$B = 100 \times 10^{-6}\; \sin\!\left[\,2\pi \times 2 \times 10^{15}\left(t-\dfrac{x}{c}\right)\right]\;.$$

From this expression we can directly read the amplitude (the maximum value) of the magnetic field. The number that multiplies the sine function is the amplitude, so

$$B_0 = 100 \times 10^{-6}\ \text{tesla}.$$

Now we simplify this amplitude:

$$B_0 = 100 \times 10^{-6}\; \text{T} = 1.00 \times 10^{-4}\; \text{T}.$$

For any electromagnetic wave travelling in free space, the magnitudes of the electric and magnetic field amplitudes are related by the universally valid relation

$$\dfrac{E_0}{B_0} = c,$$

where $$c$$ is the speed of light in vacuum. This formula can be rearranged to obtain the required electric field amplitude:

$$E_0 = c\,B_0.$$

The speed of light is given in the problem as

$$c = 3 \times 10^{8}\ \text{m s}^{-1}.$$

Substituting $$B_0 = 1.00 \times 10^{-4}\; \text{T}$$ and $$c = 3 \times 10^{8}\; \text{m s}^{-1}$$ in the relation $$E_0 = c\,B_0$$, we get

$$\begin{aligned} E_0 &= (3 \times 10^{8}) \times (1.00 \times 10^{-4}) \\ &= 3 \times (10^{8} \times 10^{-4}) \\ &= 3 \times 10^{4}\ \text{V m}^{-1}. \end{aligned}$$

The SI unit $$\text{V m}^{-1}$$ is identical to $$\text{N C}^{-1}$$, so

$$E_0 = 3 \times 10^{4}\ \text{N C}^{-1}.$$

Hence, the correct answer is Option A.