Answer
${\bf16.3}^\circ\tag{CCW from $+x$-direction}$
Work Step by Step
According to the Galilean fields transformation equations,
$$\left.\begin{matrix}
&\vec E_B=\vec E_A+\vec v_{BA}\times \vec B_A \\
& \vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}\\
\end{matrix}\right\}\tag 1$$
We are given that
- $B_A=(0.5\;\hat k)\;\rm T$,
- $E_A=(10^6\cos45^\circ\;\hat i+10^6\sin45^\circ\;\hat j)\;\rm V/m$,
- $\vec v_{BA}=(10^6\hat i)\;\rm m/s$
To find $\vec E_B$, we can use the first Galilean's fields transformation equation equation above.
$$\vec E_B=\vec E_A+\vec v_{BA}\times \vec B_A $$
Plug the known;
$$\vec E_B=(10^6\cos45^\circ\;\hat i+10^6\sin45^\circ\;\hat j)+(10^6\hat i)\times (0.5\;\hat k) $$
$$\vec E_B=(10^6\cos45^\circ\;\hat i+10^6\sin45^\circ\;\hat j)- (0.5\times 10^6\;\hat j) $$
$$\vec E_B=(10^6\cos45^\circ\;\hat i+10^6\sin45^\circ\;\hat j)- (0.5\times 10^6\;\hat j) $$
$$\vec E_B=( 7.07\times 10^5 \;\hat i+2.07\times 10^5 \;\hat j) \;\rm V/m$$
So its angle is given by
$$\theta=\tan^{-1}\left[\dfrac{2.07\times 10^5}{ 7.07\times 10^5}\right]$$
$$\theta=\color{red}{\bf16.3}^\circ\tag{CCW from $+x$-direction}$$
