Answer
${\bf 0.853}\;\rm mA$
Work Step by Step
We have here a loop and a changing flux, so we must have an induced current and an induced emf.
We know that the induced current is given by
$$I=\dfrac{\varepsilon}{R}\tag 1$$
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon=\left|\dfrac{d\Phi}{dt}\right|\tag 2$$
where $ \Phi =\vec A\cdot \vec B =AB\cos\theta=AB $ since $\theta=0^\circ$, so
$$d\Phi = BdA=0.8 y^2 t L dy$$
Integrating,
$$\int_0^\Phi d\Phi=0.8tL\int_0^L y^2 dy$$
$$ \Phi=\dfrac{0.8tL }{3} y^3\bigg|_0^L $$
$$ \Phi=\dfrac{0.8tL^4 }{3} $$
Plug into (2),
$$\varepsilon=\left|\dfrac{d }{dt}\dfrac{0.8tL^4 }{3} \right| $$
$$\varepsilon=\left| \dfrac{0.8 L^4 }{3} \right| $$
Plug into (1),
$$I=\dfrac{0.8 L^4}{3R}$$
It is obvious now that the induced current is constant.
Plug the known;
$$I=\dfrac{0.8 (0.2)^4}{3(0.5)}$$
$$I=\color{red}{\bf 0.853}\;\rm mA$$
