A plane electromagnetic wave with a frequency of 30 MHz travels in free space. At a particular point in space and time, the electric field is 6 V m$$^{-1}$$. The magnetic field at this point will be $$x \times 10^{-8}$$ T. The value of $$x$$ is _________.

We have a plane electromagnetic wave propagating in free space. In such a wave, the electric field $$\mathbf E$$ and magnetic field $$\mathbf B$$ are always related by the fundamental relation for free space

$$E = c\,B,$$

where $$E$$ is the instantaneous (or peak) value of the electric field, $$B$$ is the corresponding value of the magnetic field, and $$c$$ is the speed of light in vacuum. The universally accepted value of the speed of light is

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

At the given point, the electric field is reported as

$$E = 6\ \text{V m}^{-1}.$$

Substituting this value and the value of $$c$$ into the relation $$E = c\,B$$, we can solve for the magnetic field $$B$$:

$$B = \dfrac{E}{c}.$$

Now inserting the numerical values, we get

$$B = \dfrac{6\ \text{V m}^{-1}}{3 \times 10^{8}\ \text{m s}^{-1}}.$$

Performing the division step by step:

First divide the numeric coefficients: $$6 \div 3 = 2.$$

Next, divide the powers of ten: $$10^{0} \div 10^{8} = 10^{-8}.$$

Putting these together, we obtain

$$B = 2 \times 10^{-8}\ \text{T}.$$

The problem statement expresses the magnetic field in the form $$x \times 10^{-8}\ \text{T}$$, so by direct comparison we identify

$$x = 2.$$

So, the answer is $$2$$.