The magnetic field of a plane electromagnetic wave in vacuum is...

The phase of the wave is $(kx + \omega t)$. Since both the $x$ term and the $t$ term have the same positive sign, the wave is propagating in the negative $x$-direction ($-\hat{i}$).

The direction of propagation is given by the cross product $\vec{E} \times \vec{B}$.
We know the magnetic field $\vec{B}$ oscillates along the $+\hat{j}$ direction.
Let the direction of the electric field be $\hat{n}$.
$\hat{n} \times \hat{j} = -\hat{i}$
Since $\hat{k} \times \hat{j} = -\hat{i}$, the electric field must oscillate along the $+\hat{k}$ direction.

The amplitude of the electric field $E_0$ is related to the magnetic field amplitude $B_0$ by:
$E_0 = c B_0 = (3 \times 10^8 \text{ m/s}) \times (2 \times 10^{-8} \text{ T}) = 6 \text{ V/m}$

Combining the amplitude, phase, and direction, the electric field equation is:
$\vec{E} = 6 \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t) \hat{k} \text{ V/m}$

Answer: $6 \sin(0.5 \times 10^3 x + 1.5 \times 10^{11} t) \hat{k}$ V/m