Everything you need to solve this problem has arisen in the other capacitor problems you have worked on recently.

The starting point is to look for formulae which connect the separation of the plates $x$, the capacitance $C$, the charge stored $Q$, and the potential difference $V$. Note that (i) the total charge $Q_1+Q_2$ on the 2 capacitors is fixed, and (ii) the PD across the capacitors is always the same : $V_1=V_2$.

The charge on $C_1$ will **increase**, not decrease : the plates are getting closer so the capacitance increases and the charge increases with it. *Vice versa* for $C_2$.

The speed of the plates is $\frac{dx}{dt}$. The current leaving $C_2$ and arriving on $C_1$ is $\frac{dQ_1}{dt}=-\frac{dQ_2}{dt}$ (because $Q_1$ is increasing).

The total charge on the two capacitors is fixed : $$Q_1+Q_2=Q_0=400\mu C$$ The capacitance of each capacitor is $C=\frac{\epsilon A}{x}$. The voltage across each capacitor changes as the separation between the plates changes, but the two voltages are always equal so $$\frac{Q_1}{C_1}=\frac{Q_2}{C_2}$$ $$Q_1 x_1=Q_2 x_2$$ $$\dot Q_1x_1+Q_1\dot x_1=\dot Q_2x_2+Q_2\dot x_2$$ $$\dot Q_1(x_1+x_2)=-\dot x_1(Q_2+Q_1)=-\dot x_1 Q_0$$ in which I have used the fact that $\dot Q_2=-\dot Q_1$ and $Q_2=Q_0-Q_1$ and $\dot x_2=-\dot x_1$.

It is not clear at what *moment* the current in the circuit is to be evaluated. I shall assume it is $t=0$. Then $x_1+x_2=0.2m$ and $\dot x_1=-0.001 m/s$ so the current in the circuit is $$\dot Q_1(0)=\frac{0.001m/s}{0.2m}\times 400\mu C=2.0A$$

In fact $x_1+x_2=\text{constant}$ so this is the constant current which flows until the plates of $C_2$ touch at $t=100s$ when $C_1$ discharges because there is a short-circuit across it.