For the reaction $$2\text{NO}_{2(g)} \rightleftharpoons 2\text{NO}_{(g)} + \text{O}_{2(g)}$$, $$(K_c = 1.8 \times 10^{-6} \text{ at } 184^\circ\text{C})$$, $$(R = 0.0831 \text{ kJ}/(\text{mol} \cdot \text{K}))$$. When $$K_p$$ and $$K_c$$ are compared at $$184^\circ$$C, it is found that

For a gaseous equilibrium the pressure-form and concentration-form constants are related by

$$K_p = K_c\,(RT)^{\Delta n}$$

where $$\Delta n$$ is the difference between total moles of gaseous products and gaseous reactants.

For the reaction
$$2\,\text{NO}_{2(g)} \rightleftharpoons 2\,\text{NO}_{(g)} + \text{O}_{2(g)}$$

$$\Delta n = (2 + 1) - 2 = 1$$

The temperature is $$184^{\circ}\text{C} = 184 + 273 = 457\text{ K}$$.
Using the given gas constant $$R = 0.0831 \text{ kJ}\,(\text{mol}^{-1}\,\text{K}^{-1})$$,

$$RT = 0.0831 \times 457 \approx 38 \text{ kJ mol}^{-1}$$

Since $$RT \gt 1$$ and $$\Delta n = 1$$, we have

$$K_p = K_c\,(RT)^{1} = K_c \times (RT) \gt K_c$$

Thus $$K_p$$ is numerically larger than $$K_c$$ at $$184^{\circ}\text{C}$$.

Option A which is: $$K_p$$ is greater than $$K_c$$