Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The wire reaches the terminal speed when the net force exerted on it in the $y$-direction is zero.
We have here two forces exerted on the sliding wire, the gravitational force (downward) and the magnetic force and we need to find the direction of this force.
The wire initially slides down, so the area of the loop increases which means an increase in the flux during the sliding down.
So, according to Lenz's law, the induced current due to this change in the net flux will fight this change.
Hence, the induced flux must be out of the page which means that the induced current through the loop is clockwise.
This means that the direction of the current on the sliding wire is from left to right. According to the right-hand rule, this magnetic force must be upward.
Applying Newton's second law,
$$\sum F_y=F_B-mg=ma_y=m(0)=0$$
Thus,
$$ BIl=mg\tag1 $$
where $I$ is the induced current which is given by
$$I_{\rm loop}=\dfrac{\varepsilon_{\rm loop }}{R_{\rm wire}}\tag 2$$
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon_{\rm loop}=\left|\dfrac{d\Phi_{\rm }}{dt}\right|\tag 3$$
where $ \Phi =\vec A\cdot \vec B =AB\cos90^\circ=AB $
where $A=ly$
So,
$$ \Phi = lyB $$
Plug into (3);
$$\varepsilon_{\rm loop}=\left|\dfrac{d }{dt}lyB \right| $$
$$\varepsilon_{\rm loop}=lB\cos\theta\left|\dfrac{dy }{dt}\right| $$
$$\varepsilon_{\rm loop}=lBv_y=lBv_{\rm term} $$
Plug into (2),
$$ I_{\rm loop}=\dfrac{ lBv_{\rm term} }{R} $$
Plug into (1),
$$ \dfrac{ B^2l^2v_{\rm term} }{R} =mg $$
Hence,
$$\boxed{v_{\rm term} =\dfrac{mgR}{B^2l^2} }$$
$$\color{blue}{\bf [b]}$$
Plug the given into the boxed formula above,
$$v_{\rm term} =\dfrac{(10\times 10^{-3})(9.8)(0.1)}{(0.5)^2(0.2)^2}$$
$$v_{\rm term} =\color{red}{\bf 0.98}\;\rm m/s $$
