Answer
See the detailed answer below.
Work Step by Step
We have here a loop and a changing flux, so we must have an induced current.
We know that the induced current is given by
$$I=\dfrac{\varepsilon}{R}\tag 1$$
where the induced emf $\varepsilon$ is given by
$$\varepsilon=\left|\dfrac{d\Phi}{dt}\right|$$
where $\phi =AB\cos\theta$ and here $A$ is constant and $\theta= 0^\circ$, so that
$$\varepsilon=A\left|\dfrac{dB}{dt}\right|$$
where $B$ is given by $B=4t-2t^2$;
$$\varepsilon= A\left|\dfrac{d }{dt}(4t-2t^2)\right|$$
$$\varepsilon= L^2\left (4 -4t \right)\tag 2$$
where $L$ is the side length of the square.
Plug $\varepsilon$ from (2) into (1),
$$I=\dfrac{ L^2}{R} \left(4 -4t \right)$$
Plug the known;
$$I=\dfrac{ 0.20^2}{0.10} \left(4 -4t \right)$$
$$I=1.6\left(1 -t \right)$$
At $t=0$ s,
$$I=1.6\left(1 -0\right)$$
$$I=\color{red}{\bf 1.6}\;\rm A$$
At $t=1$ s,
$$I=1.6\left(1 -1\right)$$
$$I=\color{red}{\bf 0}\;\rm A$$
At $t=2$ s,
$$I=1.6\left(1 -2\right)$$
$$I=\color{red}{\bf -1.6}\;\rm A$$
The negative sign indicates an opposite direction of the current.
