An electromagnetic wave travels in free space along the x-direction. At a particular point in space and time, $$\vec{B} = 2 \times 10^{-7}\hat{j}$$ T is associated with this wave. The value of corresponding electric field $$\vec{E}$$ at this point is _________ V/m.

The wave is moving in the $$+x$$-direction, so for a plane electromagnetic (EM) wave in free space the three vectors $$\vec{E}, \vec{B}, \vec{k}$$ are mutually perpendicular and obey the right-hand rule

$$\vec{E}\times\vec{B} \;=\; \dfrac{1}{\mu_0}\,(\text{Poynting vector}) \propto \vec{k},$$
therefore $$\vec{E}\times\vec{B}$$ must point in the $$+x$$-direction.

Given $$\vec{B}=2\times10^{-7}\,\hat{j}\; \text{T}$$ (along $$+y$$), let the unknown electric field be $$\vec{E}=E_z\,\hat{k}$$ (along $$\pm z$$). Compute the cross product:

$$\vec{E}\times\vec{B}=E_z\,\hat{k}\;\times\;2\times10^{-7}\,\hat{j}$$

Use the cyclic relation $$\hat{i}\times\hat{j}=\hat{k}, \;\hat{j}\times\hat{k}=\hat{i}, \;\hat{k}\times\hat{i}=\hat{j}$$, so

$$\hat{k}\times\hat{j}=-\hat{i}.$$

Hence

$$\vec{E}\times\vec{B}=E_z\;(2\times10^{-7})\,(-\hat{i}).$$

To make this result point along $$+\hat{i}$$ (the propagation direction), we need $$E_z$$ to be negative. Therefore the electric field must point along $$-\hat{k}$$.

In free space the magnitudes satisfy $$|\vec{E}| = c\,|\vec{B}|$$, where $$c = 3\times10^{8}\,\text{m/s}$$. Thus

$$|\vec{E}| = c\,|\vec{B}| = (3\times10^{8})(2\times10^{-7}) = 6.0\times10^{1} \;\text{V/m} = 60 \;\text{V/m}.$$

Adding the direction found earlier,

$$\boxed{\;\vec{E} = -60\,\hat{k}\;\text{V/m}\;}.$$

Hence the correct option is:
Option B which is: $$-60\hat{k}$$ V/m