To determine the direction in which the reaction will proceed, we need to compare the reaction quotient $Q_c$ with the equilibrium constant $K_c. \\$The given reaction is:$\\ 2A \rightleftharpoons B + C \\$The equilibrium constant $K_c$ is given as $4 \times 10^{-3}. \\$At a given time, the concentrations are:$\\ [A] = 2 \times 10^{-3} \,$M$, \\ [B] = 2 \times 10^{-3} \,$M$, \\ [C] = 2 \times 10^{-3} \,$M$. \\$The reaction quotient $Q_c$ is calculated as follows:$\\ Q_c = \frac{[B][C]}{[A]^2} \\$Substitute the given concentrations into the expression for $Q_c: \\ Q_c = \frac{(2 \times 10^{-3})(2 \times 10^{-3})}{(2 \times 10^{-3})^2} \\ Q_c = \frac{4 \times 10^{-6}}{4 \times 10^{-6}} \\ Q_c = 1 \\$Now, compare $Q_c$ with $K_c: \\ Q_c = 1 \quad$and$\quad K_c = 4 \times 10^{-3} \\$Since $Q_c > K_c,$ the reaction will proceed in the backward direction to reach equilibrium.$\\$Therefore, the correct option is:$\\$Option 1: Reaction has a tendency to go in backward direction.