Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The maximum current occurs at the moment the two plates are connected to each other by the copper wire. Hence just after the connection, the voltage is maximum which is the voltage of parallel plate capacitors.
Thus,
$$\Delta V_{max}=\Delta V_C=I_{max} R_{wire}$$
Hence,
$$I_{max}=\dfrac{\Delta V_C}{R_{wire}}$$
where $R=\rho_{\rm copper} L/A_{wire}$, and $\Delta V_C=Q/C$
$$I_{max}=\dfrac{Q A_{wire}}{ \rho_{\rm copper} CL}$$
where $C=\epsilon_0A_{plate}/d$;
$$I_{max}=\dfrac{Q A_{wire }d }{A_{plate}\epsilon_0 \rho_{\rm copper} L}$$
where $A=\pi r^2$, so
$$I_{max}=\dfrac{\color{red}{\bf\not} \pi r_{wire}^2Q \color{red}{\bf\not}d }{\color{red}{\bf\not} \pi r_{plate}^2\epsilon_0 \rho_{\rm copper} \color{red}{\bf\not}L}$$
where $L=d$ since the wire is stretched between the two plates from their centers.
$$I_{max}=\dfrac{ r_{wire}^2Q }{ r_{plate}^2\epsilon_0 \rho_{\rm copper} }$$
Plug the known;
$$I_{max}=\dfrac{(0.112\times 10^{-3})^2(12.5\times 10^{-9}) }{ (0.05)^2(8.85\times 10^{-12})(1.7\times 10^{-8}) }$$
$$I_{max}=\color{red}{\bf 4.17\times 10^5}\;\rm A$$
$$\color{blue}{\bf [b]}$$
The current decreases with time until it becomes zero when the two plates are then neutral.
So the answer here is $\bf decreased$.
$$\color{blue}{\bf [c]}$$
Since the energy is conserved and our system is isolated, the energy stored in the capacitor before connecting its plates together is the same amount of energy dissipated in the wire
$$E_{\rm dissipated }=U_C=\dfrac{Q^2}{2C}$$
where $C=\epsilon_0A_{plate}/d$;
$$E_{\rm dissipated }=\dfrac{Q^2d}{2\epsilon_0A_{plate}}$$
$$E_{\rm dissipated }=\dfrac{Q^2d}{2\pi \epsilon_0r_{plate}^2}$$
Plug the known;
$$E_{\rm dissipated }=\dfrac{(12.5\times 10^{-9})^2(0.01)}{2\pi (8.85\times 10^{-12})(0.05)^2 }$$
$$E_{\rm dissipated }=\color{red}{\bf 1.124\times 10^{-5}}\;\rm J$$
