Answer
${\bf 1.0}\;\rm mA$
Work Step by Step
We have here a loop and a changing flux due to the moving away from the current carrying 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 can see that the flux through the loop decreases since the loop moves away from the wire. According to Lenz’s law, the induced
current must be clockwise to increase the flux (or to maintain the original net flux).
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon_{\rm loop}=\left|\dfrac{d\Phi_{\rm wire}}{dt}\right|=\left|\dfrac{d\Phi_{\rm wire}}{dh}\cdot \dfrac{dh}{dt}\right|$$
where $h$ is the distance from the wire to the closer edge of the loop, and hence $dh/dt=v$;
$$\varepsilon_{\rm loop}= \left|\dfrac{vd\Phi_{\rm wire}}{dh} \right|\tag 2$$
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,
$$\int_0^\Phi d\Phi= \int_h^{h+W} \dfrac{\mu_0 IL dy}{2\pi y} $$
$$ \Phi= \dfrac{\mu_0 IL }{2\pi } \int_h^{h+W} \dfrac{ dy}{ y} $$
$$ \Phi= \dfrac{\mu_0 IL }{2\pi }\ln y \bigg|_h^{h+W} $$
$$ \Phi= \dfrac{\mu_0 IL }{2\pi }\ln\left[\dfrac{h+W}{h}\right] $$
Plug into (2),
$$\varepsilon_{\rm loop}= \left|\dfrac{\mu_0 ILv }{2\pi }\dfrac{ d }{dh}\ln\left[\dfrac{h+W}{h}\right]\right| $$
$$\varepsilon_{\rm loop}= \left|\dfrac{\mu_0 ILv }{2\pi }\dfrac{ d }{dh}\left(\ln\left[{h+W}\right]-\ln \left[{h}\right]\right)\right| $$
$$\varepsilon_{\rm loop}= \left|\dfrac{\mu_0 ILv }{2\pi } \left(\dfrac{1}{h+W} -\dfrac{1}{h} \right)\right| $$
$$\varepsilon_{\rm loop}= \left|\dfrac{\mu_0 ILv }{2\pi } \left(\dfrac{1}{h+W} -\dfrac{1}{h} \right)\right| $$
Plug into (1),
$$I_{\rm loop}=\dfrac{ \left|\dfrac{\mu_0 ILv }{2\pi } \left(\dfrac{1}{h+W} -\dfrac{1}{h} \right)\right|}{R} $$
Plug the known;
$$I_{\rm loop}=\dfrac{ \left|\dfrac{(4\pi \times 10^{-7}) (15)(0.04)(10) }{2\pi } \left(\dfrac{1}{h+0.01} -\dfrac{1}{h} \right)\right|}{0.02} $$
At this instant, $h=0.02$ m,
$$I_{\rm loop}=\dfrac{ \left|\dfrac{(4\pi \times 10^{-7}) (15)(0.04)(10) }{2\pi } \left(\dfrac{1}{0.02+0.01} -\dfrac{1}{0.02} \right)\right|}{0.02} $$
$$I_{\rm loop}=\color{red}{\bf 1.0}\;\rm mA$$
