The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is $$\vec{E} = E_0 \hat{j}\cos(\omega t - kx)$$. The magnetic field $$\vec{B}$$, at the moment t = 0 is:

We are told that the electromagnetic (EM) wave is travelling in vacuum along the $$+x$$-direction and that its electric field is

At the instant $$t=0$$ this reduces to

For a plane EM wave in free space we always have three basic facts:

(i) The electric field $$\vec E$$, the magnetic field $$\vec B$$, and the propagation vector $$\vec k$$ are mutually perpendicular.

(ii) The magnitudes obey the relation $$E_0=cB_0,$$ where $$c=\frac1{\sqrt{\mu_0\varepsilon_0}}$$ is the speed of light in vacuum.

(iii) The vectors follow the right-hand rule: $$\vec E\times\vec B$$ points in the direction of $$\vec k$$.

We already know $$\vec k$$ points along $$+x$$, so $$\hat k\_x=\hat{\imath}$$. The given $$\vec E$$ points along $$+\hat{\jmath}$$, that is the $$+y$$-axis. To satisfy perpendicularity and the right-hand rule, $$\vec B$$ must point along $$+\hat{k}$$, i.e. the $$+z$$-axis, because

Now we find the amplitude of $$\vec B$$. From (ii) we write

The space dependence of the fields in a monochromatic plane wave is the same for $$\vec E$$ and $$\vec B$$. Therefore, at $$t=0$$ the magnetic field should also carry the factor $$\cos(kx)$$.

Putting magnitude, direction and spatial variation together, we obtain

This matches exactly Option C.

Hence, the correct answer is Option 3.