Answer
See the detailed answer below.
Work Step by Step
We have here a loop and a changing flux due to the changing in the current in the solenoid, so we must have an induced current and an induced emf.
We know that the induced current is given by
$$I_{\rm loop}=\dfrac{\varepsilon_{\rm loop}}{R}\tag 1$$
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon_{\rm loop}=\left|\dfrac{d\Phi_{\rm solenoid}}{dt}\right|\tag 2$$
where $ \Phi =\vec A\cdot \vec B =AB\cos\theta=AB $ since $\theta=0^\circ$, so
$$d\Phi =A dB$$
where $B_{\rm solenoid}=\mu_0 NI/l$
$$d\Phi =A_{\rm solenoid}\dfrac{\mu_0NdI}{l} =\pi r^2_{\rm solenoid}\dfrac{\mu_0NdI}{l} $$
We used the area of the solenoid not that of the loop since we assumed that the solenoid is an ideal one and the flux outside its area is zero.
Plug into (2),
$$\varepsilon_{\rm loop}=\left|\pi r^2_{\rm solenoid}\dfrac{\mu_0N}{l} \dfrac{dI }{dt}\right| $$
$$\varepsilon_{\rm loop}=\pi r^2_{\rm solenoid}\dfrac{\mu_0N}{l} \left|\dfrac{dI }{dt}\right| $$
Plug into (1),
$$I_{\rm loop}=\dfrac{\pi r^2_{\rm solenoid}\dfrac{\mu_0N}{l} \left|\dfrac{dI }{dt}\right|}{R} $$
$$\color{blue}{\bf [a]}$$
During $0\lt t\lt 1$, the current is constant, as seen from the given graph.
So that, at $t=0.5$, $dI/dt=0$, thus,
$$I_{\rm loop}=\color{red}{\bf 0}\;\rm A$$
$$\color{blue}{\bf [b]}$$
During $1\lt t\lt 2$, $|dI/dt|=40$, thus, at $t=1.5$ s,
$$I_{\rm loop}=\dfrac{\pi (0.01)^2 \dfrac{(4\pi\times 10^{-7})(100)}{(0.1)} \left|-40\right|}{(0.10)} $$
$$I_{\rm loop}=\color{red}{\bf 1.58\times 10^{-4}}\;\rm A$$
$$\color{blue}{\bf [c]}$$
During $2\lt t\lt 3$, the current is constant, as seen from the given graph.
So that, at $t=2.5$, $dI/dt=0$, thus,
$$I_{\rm loop}=\color{red}{\bf 0}\;\rm A$$