Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We have here a loop and a changing flux due to the change in the area that enters the uniform flux, so we must have an induced current and an induced emf.
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\cos\theta=AB $ since $\theta=0^\circ$, so for
$$d\Phi =dA B$$
Plug into (2),
$$\varepsilon_{\rm loop}=B\left|\dfrac{dA}{dt}\right| \tag 3$$
where $dA$ is not changing uniformly since the loop enters the flux from the corner.
At $t=0$, the right corner of the loop is at the edge of the flux area and after a short time $t$ this corner would be moved a distance of $x=vt$.
Thus the area of the loop when it is moving inside the flux zone until half of the loop's area is inside the flux zone.
$$A = \frac{1}{2}x\left(2y\right)=xy$$
and according to the equal sides length of the square loop, then $x=y$ when it is fully inside the flux zone.
$$A = xy=x^2$$
where $x=vt$ and $v$ is constant.
$$A =(vt)^2$$
Plug into (3),
$$\varepsilon_{\rm loop}=B\left|\dfrac{dv^2t^2}{dt}\right| $$
$$\varepsilon_{\rm loop}=2B v^2t $$
Plug into (1),
$$I_{\rm loop}=\dfrac{2B v^2t }{R_{\rm loop}} $$
Plug the known;
$$I_{\rm loop}=\left(\dfrac{2(0.8) (10)^2 }{(0.10)}\right)t $$
$$ \boxed{I_{\rm loop}=1600 t} $$
This is for the right side of the loop, so we can find the whole time needed for the whole loop to enter the flux zone.
where $t =2t_1$ where $t_1=s/v$, and $s=\frac{1}{2}x=\sqrt{10^2+10^2}=10\sqrt{2}\;\rm cm$.
Hence,
$$t=2t_1=\dfrac{10\sqrt{2}\times 10^{-2}}{10}=\bf 14\;\rm ms$$
So, $$t_1=\bf 7\;\rm ms$$
Hence,
$$I_1=1600\times 7\times 10^{-3}=\color{blue}{\bf 11.2}\;\rm A$$
While the left side enters the flux zone, the area exposed to the flux increases but at a smaller rate, and hence the current decreases at the same rate as it increases, as we found above (the boxed formula), until it reaches zero when the whole loop is inside the flux zone.
Hence,
$$I_2=\color{blue}{\bf 0}\;\rm A$$
See the graph below.
$$\color{blue}{\bf [b]}$$
As shown in the figure below, the maximum current occurs at $t=7$ ms when the right half of the loop is inside the flux zone.
$$I_{\rm max}=\color{red}{\bf 11.2}\;\rm A$$
