For a plane electromagnetic wave, the magnetic field at a point $x$ and time $t$ is
$$\vec B(x, t) = \left[1.2\times10^{-7}\sin\left(0.5\times10^3x + 1.5\times10^{11}t\right)\hat k\right]\ \text{T}$$
The instantaneous electric field $\vec E$ corresponding to $\vec B$ is:
Answer: (A) $\vec E = \left[-36\sin(0.5\times10^3x + 1.5\times10^{11}t)\hat j\right]$ V/m
$E_0 = cB_0 = 3\times10^8\times1.2\times10^{-7} = 36$ V/m, with the same phase $(0.5\times10^3x + 1.5\times10^{11}t)$.
The phase has $+kx + \omega t$, so the wave travels along $-x$. The direction of propagation is that of $\vec E\times\vec B$; with $\vec B$ along $\hat k$ we need $\vec E\times\hat k \parallel -\hat i$, which holds for $\vec E$ along $-\hat j$ (since $-\hat j\times\hat k = -\hat i$).
$$\vec E = -36\sin(0.5\times10^3x + 1.5\times10^{11}t)\,\hat j\ \text{V/m}$$
Solution by Sreeraj P, M.Sc Physics