Is it possible to find the volume of condensed water by temperature change without knowing the initial pressure of the flask?
This problem is sloppily worded and, technically, it is not possible to give a definitive answer based on the information provided.  But the missing information is not what you think -- you don't need to know the air pressure in the flask.  What's missing are two key pieces:  The degree of humidity ("humid" doesn't mean "saturated") and the volume of water in the flask.  If the flask contains no water, and air that, while humid, is sufficiently below 100% $RH$ at $\pu{30 ^\circ C}$, there might be no water condensation when cooling to $\pu{27 ^\circ C}$ (i.e. you'd just be increasing the $RH$ from some number &lt;100% to a value that, while larger, is still &lt;100%). Alternately, if the flask is initially half liquid water, then the volume of water that condenses would be about half that if the flask instead initially contained a very small volume of liquid water. Thus here is what I think the author meant to ask:


  A sealed container of $\pu{6.24 L}$ contains an equilibrated mixture of air and liquid water at $\pu{30 ^\circ C}$. The container temperature is then lowered to $\pu{27 ^\circ C}$. Find the volume of water that condenses. Assume the volume of liquid water is insigificant relative to the volume of the container.  The vapor pressure of water is $\pu{32 mmHg}$ at $\pu{30 ^\circ C}$ and $\pu{27 mmHg}$ at $\pu{27 ^\circ C}$.


As to how to approach the question, that's straightforward:  You know the partial pressure of the water vapor at each temperature.  You also know the total volume of the air.  Given this, you can determine the number of moles of water vapor at each temperature (using the ideal gas law).  The difference is the number of moles of water that condenses, from which you can then determine the volume of the water that condenses.