Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The wire here is sliding down a U-shaped rail with no friction and the flux is vertical to the surface of the rail which means that the angle between the flux vector and the area vector is equal to the angle between the rail and the horizontal.
The induced current is given by
$$I_{\rm loop}=\dfrac{\varepsilon_{\rm loop }}{R_{\rm wire}}\tag 1$$
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon_{\rm loop}=\left|\dfrac{d\Phi_{\rm }}{dt}\right|\tag 2$$
where $ \Phi =\vec A\cdot \vec B =AB\cos\theta $
where $A$ changes while the wire slides down the rail. It increases if its open side is down the rail and it decreases if the open side is up the rail.
And since we don't know the direction of the open side of the rail or the direction of the magnetic field up or down, let's assume that the open side is down the rail and that the magnetic field is upward (out of the loop).
Hence the area increases since $x$ increases, as we see in the figure below.
$$A=lx$$
So,
$$ \Phi = lxB\cos\theta $$
Plug into (2);
$$\varepsilon_{\rm loop}=\left|\dfrac{d }{dt}lxB\cos\theta\right| $$
$$\varepsilon_{\rm loop}=lB\cos\theta\left|\dfrac{dx }{dt}\right| $$
$$\varepsilon_{\rm loop}=lBv_x\cos\theta $$
Plug into (1),
$$\boxed{I_{\rm loop}=\dfrac{ lBv \cos\theta }{R} }$$
$$\color{blue}{\bf [b]}$$
The wire reaches the terminal speed when the net force exerted on it in the $x$-direction is zero.
Applying Newton's second law,
$$\sum F_x=mg\sin\theta-F_B\cos\theta=ma_x=m(0)=0$$
Thus,
$$ F_B\cos\theta=mg\sin\theta$$
where $F_B=BIl$;
$$ BIl\cos\theta=mg\sin\theta$$
Plug $I$ from the boxed formula,
$$ \dfrac{ B^2l^2v (\cos\theta)^2 }{R} =mg\sin\theta$$
and since the net force is zero, so $v=v_{\rm term}$
$$ \dfrac{ B^2l^2v_{\rm term} (\cos\theta)^2 }{R} =mg\sin\theta$$
$$v_{\rm term} =\dfrac{mg R\sin\theta}{B^2l^2(\cos\theta)^2 }$$
$$\boxed{v_{\rm term} =\dfrac{mg R\tan\theta}{B^2l^2\cos\theta }}$$
