Answer
See the detailed answer below.
Work Step by Step
According to the Galilean fields transformation equations,
$$\left.\begin{matrix}
&\vec E_B=\vec E_A+\vec v_{BA}\times \vec B_A \\
& \vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}\\
\end{matrix}\right\}\tag 1$$
Let's assume that Earth is the reference frame A, and the rocket is the reference frame B.
So we are given that
- $B_B=(1.0\;\hat j)\;\rm T$,
- $E_B=( 10^6 \;\hat k)\;\rm V/m$,
- $\vec v_{BA}=(2\times 10^6\hat i)\;\rm m/s$
We need to find the electric field and the magnetic field measured from Earth, which are $E_A$, and $B_A$.
To find $\vec E_A$, we can use the first Galilean's fields transformation equation equation above.
$$\vec E_B=\vec E_A+\vec v_{BA}\times \vec B_A $$
$$ \vec E_A=\vec E_B-\vec v_{BA}\times \vec B_A \tag 1$$
Plug into the second formula from (1) above,
$$\vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times\left[\vec E_B-\vec v_{BA}\times \vec B_A\right]}{c^2}$$
Plug the known;
$$(1.0\;\hat j)=\vec B_A-\dfrac{(2\times 10^6\hat i)\times\left[( 10^6 \;\hat k)-(2\times 10^6\hat i)\times \vec B_A\right]}{c^2}$$
$$(1.0\;\hat j)=\dfrac{c^2\vec B_A-\left((2\times 10^6\hat i)\times\left[( 10^6 \;\hat k)-(2\times 10^6\hat i)\times \vec B_A\right]\right)}{c^2}$$
$$c^2(1.0\;\hat j)=c^2\vec B_A-\left((2\times 10^6\hat i)\times\left[( 10^6 \;\hat k)-(2\times 10^6\hat i)\times \vec B_A\right]\right)$$
$$c^2(1.0\;\hat j)=c^2\vec B_A-\left((-2\times 10^{12}\hat j) -0 \right)$$
$$c^2(1.0\;\hat j)=c^2\vec B_A+(2\times 10^{12}\hat j) $$
Solving for $\vec B_A$,
$$ \vec B_A =(1.0\;\hat j)-\dfrac{(2\times 10^{12}\hat j) }{c^2} $$
Plug the known;
$$ \vec B_A =(1.0\;\hat j)-\dfrac{(2\times 10^{12}\hat j) }{(3\times 10^8)^2} $$
$$ \vec B_A =(\color{red}{\bf 0.999977}\;\hat j)\;\rm T$$
Now we need to plug the known into (1),
$$ \vec E_A=( 10^6 \;\hat k)- (2\times 10^6\hat i)\times (0.999977\;\hat j) $$
$$ \vec E_A=( 10^6 \;\hat k)- (1.999954\times 10^6\hat k) $$
$$ \vec E_A= (\color{red}{\bf -0.999954\times 10^6}\hat k)\;\rm V/m $$
