For the reaction,
$$A(g) + B(g) \to C(g) + D(g)$$, $$\Delta H°$$ and $$\Delta S°$$ are, respectively, $$-29.8$$ kJ mol$$^{-1}$$ and $$-0.100$$ kJ K$$^{-1}$$ mol$$^{-1}$$ at 298 K. The equilibrium constant for the reaction at 298 K is:

For the gaseous reaction $$A(g)+B(g)\rightarrow C(g)+D(g)$$ we are given the standard enthalpy change $$\Delta H^\circ=-29.8\;\text{kJ mol}^{-1}$$ and the standard entropy change $$\Delta S^\circ=-0.100\;\text{kJ K}^{-1}\text{ mol}^{-1}$$ at the temperature $$T=298\;\text{K}$$.

First we recall the thermodynamic relation that links the standard Gibbs free-energy change with enthalpy and entropy:

$$\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ.$$

Substituting the numerical values (and keeping every quantity in kilojoules):

$$\Delta G^\circ=\left(-29.8\;\text{kJ mol}^{-1}\right)-\left(298\;\text{K}\right)\left(-0.100\;\text{kJ K}^{-1}\text{ mol}^{-1}\right).$$

The product $$T\Delta S^\circ$$ is

$$T\Delta S^\circ=298\;\text{K}\times(-0.100\;\text{kJ K}^{-1}\text{ mol}^{-1})=-29.8\;\text{kJ mol}^{-1}.$$

So we have

$$\Delta G^\circ=-29.8\;\text{kJ mol}^{-1}-(-29.8\;\text{kJ mol}^{-1})=0\;\text{kJ mol}^{-1}.$$

Now we connect the Gibbs free-energy change with the equilibrium constant through the formula

$$\Delta G^\circ=-RT\ln K,$$

where $$R=8.314\;\text{J K}^{-1}\text{ mol}^{-1}=0.008314\;\text{kJ K}^{-1}\text{ mol}^{-1}$$ and $$T=298\;\text{K}$$.

Since $$\Delta G^\circ=0$$, we get

$$0=-RT\ln K\quad\Rightarrow\quad\ln K=0.$$

The exponential of zero is one, therefore

$$K=e^{0}=1.$$

Hence, the correct answer is Option 3.