Find ratio of quantities of water and of argon in airtight container
You should use the ideal gas law to convert between $n_i$ (amount of gas) and $p/T$. For each temperature and pressure pair you can write
$$\begin{align}n_{g,1}=\frac{p_1V}{RT_1}\\n_{g,2}=\frac{p_2V}{RT_2}\end{align}$$
where
$$\begin{align}n_{g,1}&amp;=n_{g,H_2O}+n_{Ar}\\n_{H_2O}&amp;=n_{g,H_2O}+n_{l,H_2O}\\n_{g,2}&amp;=n_{H_2O}+n_{Ar}\end{align}$$
You also know that for each individual gas the partial pressure is proportional to the number of moles (Dalton's law):
$$n_i=\frac{p_iV}{RT}$$
 The "trick" is that at $T_1=\pu{373K}$
$$p_{Ar} = p_1 - p_V$$
where $p_V$ is given as the partial pressure of water at that temperature.
You should be able to show that 
$$\frac{n_{H_2O}}{n_{Ar}}=\frac{n_{g,2}}{n_{Ar}}-1$$
and making appropriate substitutions it should not be difficult to arrive at the answer.