Answer
a) In the negative $x$-direction.
b) In the $y$-direction.
c) In the positive $x$-direction.
Work Step by Step
According to the Galilean fields transformation equations,
$$ \vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}\tag 1 $$
From the given graph, we can see that the direction of the magnetic field of frame A is in the positive $z$-direction while the electric field is in the positive $y$-direction.
Hence, $B_A=B\;\hat k$ and $E_A=E\;\hat j$
$$\color{blue}{\bf [a]}$$
From (1), to have $B_B\gt B_A$,
$$\vec B_B-\vec B_A=-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}$$
So, we need to make $\vec B_B-\vec B_A\gt 0$, and hence,
$-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}\gt 0$.
This means that $(\vec v_{BA}\times \vec E_A)$ must be in the negative $z$-direction.
But we know that $E_A$ is in the positive $y$-direction, so $v_{BA}$ must be in the negative $x$-direction.
Therefore the rocket is, in this case, moving to the left in the negative $x$-direction.
$$\color{blue}{\bf [b]}$$
From (1), to have $B_B=B_A$,
$$\vec B_B-\vec B_A=-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}=0$$
So, we need to make $\vec B_B-\vec B_A= 0$, and hence,
$-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}=0$.
This means that $(\vec v_{BA}$ and $ \vec E_A)$ must be in the same direction.
But we know that $E_A$ is in the positive $y$-direction, so $v_{BA}$ must be in the negative or positive $y$-direction.
Therefore the rocket is, in this case, moving upward or downward in the $y$-direction.
$$\color{blue}{\bf [c]}$$
From (1), to have $B_B\lt B_A$,
$$\vec B_B-\vec B_A= -\dfrac{\vec v_{BA}\times \vec E_A}{c^2} $$
So, we need to make $\vec B_B-\vec B_A\lt 0$, and hence,
$-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}\lt 0$.
This means that $(\vec v_{BA}\times \vec E_A)$ must be in the positive $z$-direction.
But we know that $E_A$ is in the positive $y$-direction, so $v_{BA}$ must be in the positive $x$-direction.
Therefore the rocket is, in this case, moving to the right in the positive $x$-direction.
