Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We need to use Galilean fields transformation equations,
$$\left.\begin{matrix}
&\vec E_B=\vec E_A+\vec v_{BA}\times \vec B_A \\
& \vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}\\
\end{matrix}\right\}\tag 1$$
Assuming that A is the wire's frame, and B is the loop's frame.
Now we need to find $B_A$ and $E_A$, so we can use (1) to find $B_B$ and $E_B$.
Recalling that the magnetic field for a very long current-carrying wire is given by
$$\boxed{\vec E_A=\dfrac{1}{4\pi\epsilon_0}\dfrac{2\lambda}{ r}\hat r}\tag 2$$
where $\hat r$ means radially away from the wire since it is positively charged.
And
$$\boxed{\vec B_A=0\;\rm T}\tag 3$$
since the wire is stationary and the charges are stationary too (there are no moving charges here).
$$\color{blue}{\bf [b]}$$
Now we can use the first formula from (1) to find $E_B$;
$$ \vec E_B=\vec E_A+\vec v_{BA}\times \vec B_A $$
where $\vec B_A=0$, so
$$ \vec E_B=\vec E_A+\vec v_{BA}\times 0=\vec E_A $$
Hence,
$$\boxed{\vec E_B=\dfrac{1}{4\pi\epsilon_0}\dfrac{2\lambda}{ r} } $$
Now we can use the second formula from (1) to find $B_B$;
$$ \vec B_B=\vec B_A-\dfrac{\vec v_{BA}\times \vec E_A}{c^2} =0-\dfrac{\vec v_{BA}\times \vec E_A}{c^2}$$
$$ \vec B_B=-\dfrac{\vec v_{BA}\times \vec E_A}{c^2} $$
Plug the known, where $v_{BA}=v\;\hat i$, and $E_A$ is from (2),
$$ \vec B_B=-\dfrac{1}{c^2} \left[v\;\hat i\times \dfrac{1}{4\pi\epsilon_0}\dfrac{2\lambda}{ r}\hat r\right]$$
Let's take a point at the top of the wire, the direction of the electric field is upward. Hence, $\hat i\times \hat j=\hat k$
$$ \vec B_B=-\dfrac{1}{c^2} \left[v \times \dfrac{1}{4\pi\epsilon_0}\dfrac{2\lambda}{ r}\;\hat k \right]$$
$$\boxed{ \vec B_B=- \dfrac{v\lambda}{ 2\pi\epsilon_0c^2r}\;\hat k}$$
This means that the direction of the magnetic field at any point above the wire is into the page.
$$\color{blue}{\bf [c]}$$
Since the loop is moving to the right, the man on the loop sees a current that passes through the wire where the current is given by
$$I=\dfrac{dQ}{dt}\tag 4$$
And we are given that the linear charge density of the wire is $\lambda$, so $\lambda=\dfrac{dq}{dx}$ where $dx$ is a small segment of the wire.
Hence, $dq=\lambda dx$, plug that into (4)
$$I=\dfrac{\lambda dx}{dt} $$
where $dx/dt=v$
$$\boxed{I=\lambda v}$$
$$\color{blue}{\bf [d]}$$
$$B_B=\dfrac{\mu_0I}{2\pi r}$$
Plug from the previous boxed formula above,
$$B_B=\dfrac{\mu_0\lambda v}{2\pi r}\cdot\dfrac{\epsilon_0}{\epsilon_0}$$
where $\mu_0 \epsilon_0=1/c^2$
$$\boxed{B_B=-\dfrac{ \lambda v}{2 \pi \epsilon_0c^2r}\;\hat k} $$
The current due to the frame of the loop is toward the left. So at any point above the wire, the direction of the magnetic field according to the right-hand rule is into the page.
$$\color{blue}{\bf [e]}$$
You can see the two boxed formulas of parts b and d above, they are identical.
