For the reaction $2\text{A} \rightleftharpoons \text{B} + \text{C}$, the reaction quotient ($Q_c$) at the given time is calculated using the expression: $Q_c = \frac{[\text{B}][\text{C}]}{[\text{A}]^2}$

Substituting the given concentrations ($[\text{A}] = [\text{B}] = [\text{C}] = 2 \times 10^{-3}\text{ M}$): $Q_c = \frac{(2 \times 10^{-3})(2 \times 10^{-3})}{(2 \times 10^{-3})^2} = 1$

We are given the equilibrium constant $K_c = 4 \times 10^{-3}$ (or $0.004$). Comparing $Q_c$ and $K_c$, we find that $Q_c > K_c$ ($1 > 0.004$). When the reaction quotient is greater than the equilibrium constant ($Q_c > K_c$), the reaction will proceed in the reverse (backward) direction to reach equilibrium .