Answer
a) ${\bf 0}\;\rm A$
b) ${\bf 79}\;\rm \mu A$
Work Step by Step
We have here a right coil and a changing flux due to the change in the current of the left coil, so we must have an induced current and an induced emf.
We know that the induced current is given by
$$I_{\rm right}=\dfrac{\varepsilon_{\rm right}}{R}\tag 1$$
We know that the induced emf $\varepsilon$ is given by
$$\varepsilon_{\rm right}=\left|\dfrac{d\Phi_{\rm left}}{dt}\right|\tag 2$$
where $ \Phi =\vec A\cdot \vec B =AB\cos\theta=AB $ since $\theta=0^\circ$, so
$$d\Phi =A dB$$
where $B_{\rm solenoid}=\mu_0 NI/l$
$$d\Phi_{\rm left} =A_{\rm right}\dfrac{\mu_0NdI_{\rm left}}{l_{\rm left}} =\pi r^2_{\rm cylinder}\dfrac{\mu_0NdI_{\rm left}}{l_{\rm left}} $$
and for $N$ turns,
$$d\Phi_{\rm left} =N_{\rm right}A_{\rm right}\dfrac{\mu_0N_{\rm left}dI_{\rm left}}{l_{\rm left}} $$
$$d\Phi_{\rm left} =\pi N_{\rm right} r^2_{\rm cylinder}\dfrac{\mu_0N_{\rm left}dI_{\rm left}}{l_{\rm left}} $$
where $N_{\rm left}=N_{\rm right}=N$,
$$d\Phi_{\rm left} =\pi r^2_{\rm cylinder}\dfrac{\mu_0N^2d I_{\rm left}}{l_{\rm left}} $$
Plug into (2),
$$\varepsilon_{\rm right}=\pi r^2_{\rm cylinder}\dfrac{\mu_0N^2}{l_{\rm left}}\left| \dfrac{dI_{\rm left} }{dt}\right| $$
where $l_{\rm left}=N D_{\rm wire}$ where $D$ is the diameter of the wire.
$$\varepsilon_{\rm right}=\pi r^2_{\rm cylinder}\dfrac{\mu_0N^2}{ND_{\rm wire}}\left| \dfrac{dI_{\rm left} }{dt}\right| $$
Plug into (1),
$$I_{\rm right}=\dfrac{\pi r^2_{\rm cylinder}\dfrac{\mu_0N}{D_{\rm wire}}\left| \dfrac{dI_{\rm left} }{dt}\right|}{R} $$
Plug the known;
$$I_{\rm right}=\dfrac{\pi (0.01)^2\dfrac{(4\pi\times 10^{-7})(20) }{(0.001)}\left| \dfrac{dI_{\rm left} }{dt}\right|}{2} \tag 3$$
$$\color{blue}{\bf [a]}$$
During $0\lt t\lt 0.1\;\rm s$, $dI/dt=0$, so at $t=0.05$ s, the induced current must be zero.
Plug into (3),
$$I_{\rm right}=\dfrac{\pi (0.01)^2\dfrac{(4\pi\times 10^{-7})(20) }{(0.001)}\left| 0\right|}{2} $$
$$I_{\rm right}=\color{red}{\bf 0}\;\rm A$$
$$\color{blue}{\bf [b]}$$
During $0.1\lt t\lt 0.3\; $, $dI/dt=20$, so at $t=0.25$ s,
Plug into (3),
$$I_{\rm right}=\dfrac{\pi (0.01)^2\dfrac{(4\pi\times 10^{-7})(20) }{(0.001)}\left| 20\right|}{2} $$
$$I_{\rm right}=\color{red}{\bf 79}\;\rm \mu A$$
