Answer
${\bf 40}\;\rm nA$
Work Step by Step
We have here a small loop and a changing flux due to the change in the current in the bigger loop, so we must have an induced current and an induced emf.
We know that the induced current is given by
$$I_{\rm small}=\dfrac{\varepsilon_{\rm small}}{R_{\rm small}}\tag 1$$
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon_{\rm small}=\left|\dfrac{d\Phi_{\rm big}}{dt}\right|\tag 2$$
where $ \Phi =\vec A\cdot \vec B =AB\cos\theta=AB $ since $\theta=0^\circ$, so for
$$d\Phi =A_{\rm small} dB_{\rm big}$$
We used the area of the small loop area not that of the bigger loop since we need that the flux change inside the smaller loop.
where $B_{\rm loop}= \dfrac{\mu_0I_{\rm big}}{2r_{\rm big}}$
$$d\Phi_{\rm big} =\pi r_{\rm small}^2\dfrac{\mu_0dI_{\rm big}}{2r_{\rm big}}$$
Plug into (2),
$$\varepsilon_{\rm small}=\left|\dfrac{\pi r_{\rm small}^2\dfrac{\mu_0dI_{\rm big}}{2r_{\rm big}}}{dt}\right|$$
$$\varepsilon_{\rm small}=\dfrac{ \pi \mu_0 r_{\rm small}^2}{2r_{\rm big}}\left|\dfrac{dI_{\rm big}}{dt}\right|$$
Plug into (1),
$$I_{\rm small}=\dfrac{\dfrac{ \pi \mu_0 r_{\rm small}^2}{2r_{\rm big}}\left|\dfrac{dI_{\rm big}}{dt}\right|}{R_{\rm small}} $$
$$I_{\rm small}= \dfrac{ \pi \mu_0 r_{\rm small}^2}{2r_{\rm big}R_{\rm small}} \left|\dfrac{(I_2-I_1)_{\rm big}}{\Delta t}\right| $$
Plug the known;
$$I_{\rm small}= \dfrac{ \pi (4\pi\times 10^{-7}) (0.001)^2}{2(0.05) (0.02)} \left|\dfrac{(-1-1) }{0.10}\right| $$
$$I_{\rm small}\approx \color{red}{\bf 40}\;\rm nA$$
