For a reaction, $$A(g) \to A(l)$$; $$\Delta H = -3RT$$. The correct statement for the reaction is:

We are told that one mole of a gaseous substance condenses to the liquid state:

$$A(g)\;\rightarrow\;A(l)$$

and that the enthalpy change for this process is given to be

$$\Delta H = -3RT.$$

To compare $$\Delta H$$ and $$\Delta U$$, we start from the thermodynamic relation that connects the two quantities at constant pressure. The formula (valid for reactions involving ideal gases) is

$$\Delta H \;=\; \Delta U \;+\; \Delta(PV).$$

Because for an ideal gas $$PV = nRT,$$ the change in the $$PV$$ term can be rewritten in terms of the change in the number of moles of gas, which we denote by $$\Delta n_g$$:

$$\Delta(PV) \;=\; \Delta(nRT) \;=\; \Delta n_g \, RT.$$

Substituting this into the first equation gives

$$\Delta H \;=\; \Delta U + \Delta n_g \, RT.$$

Now we identify $$\Delta n_g$$ for the given reaction. Initially there is $$1$$ mole of gaseous $$A$$, while after condensation there are $$0$$ moles in the gas phase. Therefore,

$$\Delta n_g \;=\; n_{\text{final (gas)}} \;-\; n_{\text{initial (gas)}} \;=\; 0 - 1 \;=\; -1.$$

Substituting $$\Delta n_g = -1$$ into the enthalpy-internal-energy relation we obtain

$$\Delta H \;=\; \Delta U + (-1)\,RT$$

or more simply

$$\Delta H \;=\; \Delta U - RT.$$

But the problem statement already gives us the numerical form of $$\Delta H$$, namely $$\Delta H = -3RT.$$ Equating the two expressions for $$\Delta H$$ we have

$$-3RT \;=\; \Delta U - RT.$$

Adding $$RT$$ to both sides yields

$$\Delta U \;=\; -3RT + RT \;=\; -2RT.$$

Now we can compare the magnitudes of the two changes:

$$|\Delta H| \;=\; | -3RT | \;=\; 3RT,$$

$$|\Delta U| \;=\; | -2RT | \;=\; 2RT.$$

Clearly,

$$|\Delta H| \;>\; |\Delta U|.$$

This inequality exactly matches Option C.

Hence, the correct answer is Option C.