Answer:
The reaction is,
\[2A\rightleftharpoons B+C\] \[{{K}_{c}}=2\times
{{10}^{-3}}\]
\[{{Q}_{c}}=\frac{[B][C]}{{{[A]}^{2}}}=\frac{3\times
{{10}^{-4}}\times 3\times {{10}^{-4}}}{{{(3\times {{10}^{-4}})}^{2}}}=1\]
\[{{Q}_{c}}>{{K}_{c}}\]
Thus, reaction will proceed in backward direction.
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