To determine which option is correct regarding the reaction state and its direction, we need to calculate the reaction quotient $Q_c$ and compare it to the equilibrium constant $K_c$. The reaction given is:

$ 2 \mathrm{A} \rightleftharpoons \mathrm{B} + \mathrm{C} $

The equilibrium constant expression $K_c$ for this reaction is:

$ K_c = \frac{[\mathrm{B}][\mathrm{C}]}{[\mathrm{A}]^2} $

Given that $K_c = 4 \times 10^{-3}$ and the concentrations of A, B, and C at this time are each $2 \times 10^{-3}$ M, we can substitute these values into the expression for $K_c$ to calculate the reaction quotient $Q_c$:

$ Q_c = \frac{(2 \times 10^{-3} \mathrm{M})(2 \times 10^{-3} \mathrm{M})}{(2 \times 10^{-3} \mathrm{M})^2} $

Simplifying, we find:

$ Q_c = \frac{4 \times 10^{-6} \mathrm{M}^2}{4 \times 10^{-6} \mathrm{M}^2} = 1 $

Comparing $Q_c$ with $K_c$:

$ Q_c = 1 $

$ K_c = 4 \times 10^{-3} $

Since $Q_c > K_c$ (1 > 0.004), the reaction quotient is greater than the equilibrium constant. This indicates that the concentration of products (B and C) is too high relative to the concentration of reactants (A) for the system to be at equilibrium under these conditions.

This means that the reaction has a tendency to move in the backward direction to reach equilibrium, reducing the concentration of the products (B and C) and increasing the concentration of the reactant (A). Therefore, the correct answer to the given question is:

Option C: Reaction has a tendency to go in backward direction.