Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
As the loop enters the uniform magnetic flux zone, the area exposed to the flux increases which means an increase in the flux exerted on the loop.
This creates an induced current that creates a magnetic flux that opposes the original flux.
And since the original magnetic field is out of the page, then the induced flux must be into the page.
This means that the current in the loop, according to the right-hand rule, must be clockwise.
Hence, at least 3 wires of the loop experience a magnetic force. The force on the front wire is to the left while the forces on the upper and lower wire are canceling each other. The fourth wire experiences no force since it is not, yet, exposed to the magnetic flux.
We know that the induced current is given by
$$I_{\rm loop}=\dfrac{\varepsilon_{\rm loop }}{R_{\rm loop}}\tag 1$$
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\cos90^\circ=BA $ where $A$ changes while $B$ is constant.
Plug into (2),
$$\varepsilon_{\rm loop}=B \left|\dfrac{d A}{dt} \right| $$
where $A=lx$,
$$\varepsilon_{\rm loop}=Bl \left|\dfrac{d x}{dt} \right| =Blv $$
Plug into (1),
$$I_{\rm loop}=\dfrac{Blv}{R_{\rm loop}}\tag 3$$
Recalling that the magnetic force exerted on a current-carrying wire is given by
$$F_B=BIl$$
So the force on the front wire (the leading edge) is given by plugging $I$ from (3)
$$F_B=Bl\dfrac{Blv}{R_{\rm loop}}$$
$$F_B=\dfrac{B^2l^2}{R_{\rm loop}}v$$
And according to Newton's second law,
$$\sum F_x=-F_B=ma_x$$
The negative sign is due to the direction of the force that opposes the velocity direction.
Hence,
$$- \dfrac{B^2l^2}{R_{\rm loop}}v=ma$$
where $a=dv/dt$
$$- \dfrac{B^2l^2}{R_{\rm loop}}v=m \dfrac{dv}{dt}$$
Separating the time and the speed:
$$- \dfrac{B^2l^2}{mR_{\rm loop}}dt= \dfrac{dv}{v}$$
Integrating;
$$\int_0^t \dfrac{-B^2l^2}{mR_{\rm loop}}dt=\int_{v_0}^v \dfrac{dv}{v}$$
$$ \dfrac{-B^2l^2}{mR_{\rm loop}}t=\ln\left[ \dfrac{v}{v_0}\right]$$
Thus,
$$ e^{(-B^2l^2)t/(mR_{\rm loop})}= \dfrac{v}{v_0} $$
Therefore,
$$\boxed{v=v_0 e^{(-B^2l^2)t/(mR_{\rm loop})}}$$
$$\color{blue}{\bf [b]}$$
Plug the given;
$$v=10e^{(-0.1)^2(0.1)^2)t/(0.001)(0.001)}$$
$$v=10e^{-100t}$$
See the graph below.
