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
$$\varepsilon=A\left|\dfrac{dB}{dt}\right|$$
And for $N$ turns.
$$\varepsilon=NA\left|\dfrac{dB}{dt}\right|$$
where $B$ is given by $B=0.02t+0.01t^2$;
$$\varepsilon=NA\left|\dfrac{d }{dt}(0.02t+0.01t^2)\right|$$
$$\varepsilon=\pi r^2N \left(0.02 +0.02t \right)\tag 2$$
$$\color{blue}{\bf [a]}$$
Plug $\varepsilon$ from (2) into (1),
$$I=\dfrac{\pi r^2N }{R} \left(0.02 +0.02t \right)$$
Plug the known;
$$I=\dfrac{\pi (2.5\times 10^{-2})^2(20) }{(0.5)} \left(0.02 +0.02t \right)$$
$$\boxed{I={\bf1.57\times 10^{-3}}(1+t)}$$
$$\color{blue}{\bf [b]}$$
At $t=5$ s,
$$ I= 1.57\times 10^{-3} (1+5)$$
$$I=\color{red}{\bf 9.42}\;\rm mA$$
At $t=10$ s,
$$ I= 1.57\times 10^{-3} (1+10)$$
$$I=\color{red}{\bf 17.3}\;\rm mA$$
