Answer
See the detailed answer below.
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 $B_A=\bf0\;\rm T$, and $E_A={\bf 1.0\times 10^6}\;\rm V/m$
So, in a region of space at which $B=0$, using the first formula above,
$$\vec E_B=\vec E_A+\vec v_{BA}\times 0$$
$$\vec E_B=\vec E_A=\color{red}{\bf 1.0\times 10^6}\;\hat k\;\rm V/m$$
And hence, from the second formula,
$$ \vec B_B=0-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}$$
Plug the known;
$$ \vec B_B= -\dfrac{1}{(3\times 10^8)^2}(1.0\times 10^6\;\hat i)\times (1.0\times 10^6\;\hat k)$$
$$ \vec B_B= -\dfrac{1}{(3\times 10^8)^2}(-1.0\times 10^{12}\;\hat j) $$
$$ \vec B_B= \color{red}{\bf 1.11\times 10^{-5}}\;\hat j\;\rm T $$
