Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The length of one side of the loop is 10 cm. The initial separation distance between the two corners is given by the Pythagorean theorem.
$$d=\sqrt{2L^2}=\sqrt{2(10)^2}=\bf10\sqrt2\;\rm cm$$
The final separation between the two corners is given by
$$d'=2L=2(10)=\bf 20\;\rm cm$$
So that the total distance traveled by the two corners must be
$$\Delta x=d'-d=\bf 20-10\sqrt2\;\rm cm$$
Hence, the distance traveled by one corner is given by
$$\Delta x'=\dfrac{d'-d}{2}=\dfrac{ 20-10\sqrt2}{2}=\bf10-5\sqrt2\;\rm cm$$
Thus, the time it takes is given by
$$\Delta t=\dfrac{\Delta x'}{v}=\dfrac{(10-5\sqrt2)\times 10^{-2}}{0.293 }$$
$$\Delta t=\color{red}{\bf 0.1}\;\rm s$$
$$\color{blue}{\bf [b]}$$
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),
$$\varepsilon_{\rm loop}=B \left|\dfrac{d A}{dt} \right| \tag 3$$
We can divide the loop into 4 identical right-triangles.
So, while the two corners are moving horizontally, the total area of the loop is 4 times the area of one right triangle (any one of both).
Thus,
$$A=4A_{\rm triangle}=4\left(\frac{1}{2}xh\right) =2xh$$
where $h=\sqrt{L^2-x^2}$
$$A =2x\sqrt{L^2-x^2}\tag 4$$
From (3), we now have 2 variables, $x$ and $t$, so we need to use the chain rule,
$$\varepsilon_{\rm loop}=B \left|\dfrac{d A }{dx}\dfrac{d x }{dt} \right| $$
where $dx/dt=v$, plug from (4),
$$\varepsilon_{\rm loop}=B v\left|\dfrac{d }{dt} 2x\left(L^2-x^2\right)^{1/2}\right| $$
$$\varepsilon_{\rm loop}=2B v\left|\dfrac{d }{dt} x\left(L^2-x^2\right)^{1/2}\right| $$
$$\varepsilon_{\rm loop}=2B v\left| \left(L^2-x^2\right)^{1/2}+(-2x)(x)(\frac{1}{2})\left(L^2-x^2\right)^{-1/2}\right| $$
$$\varepsilon_{\rm loop}=2B v\left| \sqrt{L^2-x^2}-\dfrac{ x^2}{ \sqrt{L^2-x^2}}\right| $$
$$\varepsilon_{\rm loop}=2B v\left| \dfrac{L^2 -2 x^2}{ \sqrt{L^2-x^2}}\right| $$
where $x$ as a function of $t$ is given by $x=x_0+vt$ where $x_0=\frac{1}{2}\sqrt{2L^2}=\frac{\sqrt{2}}{2}L$. so
$x=\frac{\sqrt{2}}{2}L+vt$
$$\varepsilon_{\rm loop}=2B v\left| \dfrac{L^2 -2 \left[\frac{\sqrt{2}}{2}L+vt\right]^2}{ \sqrt{L^2- \left[\frac{\sqrt{2}}{2}L+vt\right]^2}}\right| $$
Plug into (1),
$${I_{\rm loop}=\dfrac{2B v }{R_{\rm loop}}\left| \dfrac{L^2 -2 \left[\frac{\sqrt{2}}{2}L+vt\right]^2}{ \sqrt{L^2- \left[\frac{\sqrt{2}}{2}L+vt\right]^2}}\right| }$$
Plug the known;
$$ I_{\rm loop}=\dfrac{2(0.5)(0.293) }{(0.1)}\left| \dfrac{0.1^2 -2 \left[\frac{\sqrt{2}}{2}(0.1)+(0.293)t\right]^2}{ \sqrt{(0.1)^2- \left[\frac{\sqrt{2}}{2}(0.1)+(0.293)t\right]^2}}\right| $$
$$\boxed{I_{\rm loop}=2.93\left| \dfrac{0.1^2 -2 \left[\frac{\sqrt{2}}{20} + 0.293 t\right]^2}{ \sqrt{(0.1)^2- \left[\frac{\sqrt{2}}{20} +0.293t\right]^2}}\right| }$$
$$\color{blue}{\bf [c]}$$
See the dots and the graph, in the second figure below.
