$\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{N} \mathrm{O}_{2}\right]^{2}}{\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]}$
$\mathrm{K}_{\mathrm{c}}=\frac{\left(1.2 \times 10^{-2} \mathrm{mol} / \mathrm{L}\right)^{2}}{4.8 \times 10^{-2} \mathrm{mol} / \mathrm{L}}$
$\mathrm{K}_{\mathrm{c}}=3 \times 10^{-3} \mathrm{mol} / \mathrm{L}$
Thus, the correct option is $A$
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$\begin{array}{*{20}{c}}
{\,\,\,\,\,\,\,\,O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}\,} \\
{\,\,\,\,||\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,} \\
{C{H_3} - C - C{H_2} - C - CN} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,} \\
{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{H_3}}
\end{array}$