Answer
${\bf 43.9}\;\rm \mu A$
Work Step by Step
We have here a loop and a changing flux due to the changing in the current in the wire, 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
$$\varepsilon_{\rm loop}=\left|\dfrac{d\Phi_{\rm wire}}{dt}\right|\tag 2$$
where $ \Phi =\vec A\cdot \vec B =AB\cos\theta=AB $ since $\theta=0^\circ$, so
$$d\Phi = BdA=B L dy$$
where $B_{\rm wire}=\mu_0 I/2\pi r$ where here $B_{\rm wire}=\mu_0 I/2\pi y$
$$d\Phi =\dfrac{\mu_0 IL dy}{2\pi y} $$
Integrating,
$$\int_0^\Phi d\Phi= \int_d^{d+L} \dfrac{\mu_0 IL dy}{2\pi y} $$
$$ \Phi= \dfrac{\mu_0 IL }{2\pi } \int_d^{d+L} \dfrac{ dy}{ y} $$
$$ \Phi= \dfrac{\mu_0 IL }{2\pi }\ln y \bigg|_d^{d+L} $$
$$ \Phi= \dfrac{\mu_0 IL }{2\pi }\ln\left[\dfrac{d+L}{d}\right] $$
Plug into (2),
$$\varepsilon_{\rm loop}=\left|\dfrac{d }{dt} \dfrac{\mu_0 IL }{2\pi }\ln\left[\dfrac{d+L}{d}\right]\right|$$
$$\varepsilon_{\rm loop}=\left|\dfrac{dI }{dt} \dfrac{\mu_0 L }{2\pi }\ln\left[\dfrac{d+L}{d}\right]\right| $$
Plug into (1),
$$I_{\rm loop}=\dfrac{ \left|\dfrac{dI }{dt} \dfrac{\mu_0 L }{2\pi }\ln\left[\dfrac{d+L}{d}\right]\right|}{R} $$
Plug the known;
$$I_{\rm loop}=\dfrac{ \left|(100) \dfrac{(4\pi \times 10^{-7})(0.02) }{2\pi }\ln\left[\dfrac{0.01+0.02}{0.01}\right]\right|}{0.01} $$
$$I_{\rm loop}=\color{red}{\bf 43.9}\;\rm \mu A$$
