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 $B_{\rm solenoid}=\mu_0 NI/l$
$$d\Phi =A_{\rm loop}\dfrac{\mu_0NdI}{l} =\pi r^2_{\rm loop}\dfrac{\mu_0NdI}{l} $$
Plug into (2),
$$\varepsilon_{\rm loop}=\left|\pi r^2_{\rm loop}\dfrac{\mu_0N}{l} \dfrac{dI }{dt}\right| $$
$$\varepsilon_{\rm loop}=\pi r^2_{\rm loop}\dfrac{\mu_0N}{l} \left|\dfrac{dI }{dt}\right| $$
Plug into (1),
$$I_{\rm loop}=\dfrac{\pi r^2_{\rm loop}\dfrac{\mu_0N}{l} \left|\dfrac{dI }{dt}\right|}{R} $$
During $0\lt t\lt 0.02$, $|dI/dt|=50$,
At $t=0.01$ s,
$$I_{\rm loop}=\dfrac{\pi (0.005)^2 \dfrac{(4\pi\times 10^{-7})(120)}{(0.08)} \left|-50\right|}{(0.50)} $$
$$I_{\rm loop}=\color{red}{\bf 1.48\times 10^{-5}}\;\rm A$$