A plane electromagnetic wave, has frequency of $$2.0 \times 10^{10}$$ Hz and its energy density is $$1.02 \times 10^{-8}$$ J m$$^{-3}$$ in vacuum. The amplitude of the magnetic field of the wave is close to $$\left(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \frac{Nm^2}{C^2}\right)$$ and speed of light $$= 3 \times 10^8$$ m s$$^{-1}$$.

We are told that the time-averaged energy density of the plane electromagnetic wave is $$u = 1.02 \times 10^{-8}\,\text{J m}^{-3}$$.

For a plane wave in vacuum the electric and magnetic fields are related by $$E_0 = c\,B_0$$, and the general expression for instantaneous energy density is

$$u = \frac{1}{2}\,\varepsilon_0 E^2 + \frac{1}{2\,\mu_0} B^2.$$

Because the electric and magnetic energies are equal when averaged over one cycle, the time-averaged total energy density becomes

$$u = \frac{1}{2}\,\varepsilon_0 E_0^2 = \frac{B_0^2}{2\,\mu_0}.$$

We need the amplitude $$B_0$$, so we rearrange the last relation:

$$u = \frac{B_0^2}{2\,\mu_0} \;\;\Longrightarrow\;\; B_0^2 = 2\,\mu_0\,u.$$

The permeability of free space is $$\mu_0 = 4\pi \times 10^{-7}\,\text{N A}^{-2} = 1.256 \times 10^{-6}\,\text{N A}^{-2}.$$

Substituting the numbers, we have

$$B_0^2 = 2 \times 1.256 \times 10^{-6}\,\text{N A}^{-2} \times 1.02 \times 10^{-8}\,\text{J m}^{-3}.$$

Multiplying the coefficients, $$2 \times 1.256 \times 1.02 = 2.563,$$ and adding the powers of ten, $$10^{-6}\times 10^{-8}=10^{-14},$$ so

$$B_0^2 \approx 2.563 \times 10^{-14}\,\text{T}^2.$$

Taking the square root,

$$B_0 = \sqrt{2.563 \times 10^{-14}} = \sqrt{2.563}\times 10^{-7}\,\text{T}.$$

Since $$\sqrt{2.563}\approx 1.60,$$ we get

$$B_0 \approx 1.60 \times 10^{-7}\,\text{T}.$$

Converting to nanoTesla using $$1\,\text{T} = 10^{9}\,\text{nT},$$

$$B_0 \approx 1.60 \times 10^{-7}\,\text{T} \times 10^{9}\,\frac{\text{nT}}{\text{T}} = 160\,\text{nT}.$$

Hence, the correct answer is Option B.