The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is :

We are told that in a plane electromagnetic wave the magnetic field has a peak (maximum) value $$B_0 = 20\ \text{nT}$$. Our aim is to find the corresponding peak value of the electric field $$E_0$$.

For a sinusoidal electromagnetic wave travelling in free space, the peak values of the electric and magnetic fields are related by the well-known equation for electromagnetic waves:

$$E_0 = c\,B_0$$

where $$c$$ is the speed of light in vacuum. We shall now use this relation step by step.

First, we write the numerical value of the speed of light:

$$c = 3.0 \times 10^8\ \text{m s}^{-1}$$

Next, we convert the given magnetic field from nanotesla to tesla, because the SI unit of magnetic field in the above formula is the tesla (T).

We have $$1\ \text{nT} = 10^{-9}\ \text{T}$$, so

$$B_0 = 20\ \text{nT} = 20 \times 10^{-9}\ \text{T}$$

Now we substitute the values of $$c$$ and $$B_0$$ into the relation $$E_0 = c\,B_0$$:

$$E_0 = \left(3.0 \times 10^8\ \text{m s}^{-1}\right) \times \left(20 \times 10^{-9}\ \text{T}\right)$$

To multiply the numbers conveniently, we first multiply the coefficients and then handle the powers of ten:

Coefficient part: $$3.0 \times 20 = 60$$

Power-of-ten part: $$10^8 \times 10^{-9} = 10^{8 + (-9)} = 10^{-1}$$

Putting them together gives

$$E_0 = 60 \times 10^{-1}\ \text{V m}^{-1}$$

Since $$10^{-1} = 0.1$$, we simplify:

$$E_0 = 60 \times 0.1\ \text{V m}^{-1} = 6\ \text{V m}^{-1}$$

Thus, the peak electric field strength corresponding to a peak magnetic field of 20 nT is $$6\ \text{V m}^{-1}$$.

Comparing this result with the options given, we see that $$6\ \text{V m}^{-1}$$ matches Option D.

Hence, the correct answer is Option D.