An electromagnetic wave travels in free space along the x...

The direction of propagation of an electromagnetic wave is given by the direction of $\vec{E} \times \vec{B}$.

Given that the wave travels along the $x$-direction, the direction of propagation is $\hat{i}$.

The magnetic field is given as $\vec{B} = 2 \times 10^{-7} \hat{j}$ T, so its direction is $\hat{j}$.

Let the direction of the electric field be $\hat{n}$. We must have:

$\hat{n} \times \hat{j} = \hat{i}$

Since $(-\hat{k}) \times \hat{j} = \hat{i}$, the direction of the electric field is $-\hat{k}$.

The magnitude of the electric field is related to the magnetic field by $E = cB$, where $c = 3 \times 10^8$ m/s is the speed of light in free space.

$E = (3 \times 10^8) \times (2 \times 10^{-7}) = 60$ V/m

Therefore, the electric field vector is $\vec{E} = -60 \hat{k}$ V/m.

Answer: $-60 \hat{k}$