For the reaction, $2\mathrm{NO},\left(g\right)\rightleftharpoons 2\mathrm{NO}\left(g\right)+$

${\mathrm{O}}_2\left(\mathrm{g}\right) ({\mathrm{K}}_{\mathrm{c}}=1.8\times {10}^{-6} \mathrm{at} {184}^{\mathrm{o}}\mathrm{C}, \mathrm{R}=0.0831\mathrm{kJ}/\mathrm{mol} \mathrm{K}).$ when 

${\mathrm{K}}_{\mathrm{p}}$ and ${\mathrm{K}}_{\mathrm{c}}$ are compared at $184 °\mathrm{C}$, it is found that

We know that

${\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}}{\left(\mathrm{RT}\right)}^{∆\mathrm{n}}$ .

For      $2{\mathrm{NO}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right) +{\mathrm{O}}_2\left(\mathrm{g}\right)$

        $2 \mathrm{mol} 2 \mathrm{mol} 2 \mathrm{mol}$

      $∆\mathrm{n} =(2+1)-2=1.$

$\mathrm{T} = 184 + 273 = 457 \mathrm{K}$

$\frac{K_p}{K_c}= (0.0831kJ mo{l}^{-}{K}^{-}\times 457K)\simeq 38$

Hence ${\mathrm{K}}_{\mathrm{p}}$ is greater than ${\mathrm{K}}_c$. So, the correct answer is (1).