To solve this problem, we need to analyze the circuit just after the switch is closed.$\\$Given:$\\$• Resistance of each resistor $R = 9.0 \, \Omega \\$• Inductance of each inductor $L = 2.0 \,$mH$= 2.0 \times 10^{-3} \,$H$\\$• EMF of the battery $\varepsilon = 18 \,$V$\\$Just after the switch is closed, the inductors behave as open circuits because the current through an inductor cannot change instantaneously.$\\$Therefore, the inductors will block any current flow through their branches.$\\$This simplifies the circuit to only the resistors in series with the battery.$\\$The circuit now consists of:$\\$• One resistor $R$ in series with the battery.$\\$The total resistance $R_{total}$ is simply $R = 9.0 \, \Omega. \\$Using Ohm's Law, the current $I$ through the battery is given by:$\\ I = \frac{\varepsilon}{R_{total}} \\$Substitute the known values:$\\ I = \frac{18}{9.0} \\ I = 2 \,$A$\\$However, since the inductors initially block the current, the actual current through the battery just after the switch is closed is $0 \,$A$. \\$Therefore, the correct answer is $0 \,$A$,$ corresponding to Option 3.