Calculating the mass of water removed by air conditioning condensation
Step 0) Realize that sometimes the answer given for a homework problem is wrong.  

In this case it appears that the correct answer is $\pu{6.11 kg}$ of water, not $\pu{6.00 kg}$ of water.  

The approach to the problem given below is to simply calculate the total water mass in the house before and after cooling, then subtract the latter from the former to get the mass of water removed by condensation.

Step I) Assemble parameters applicable before and after cooling:  

$R =$ gas constant $= \pu{8.31m^3~bar~mol^{-1}~K^{-1}}$  

$V = \pu{245m^3}$  

$MW_{water}=\pu{18.0g~mol^{-1}}$  

${n=\frac{PV}{RT}}$

Step II) Calculate water mass before cooling:  

$T = \pu{33^oC = 306~K}$  

$P_{saturation} = \mathrm{50.3~mbar^{(1)}}$  

$P = 80\%$ of $P_{saturation} = \mathrm{40.2~mbar}$  

${n=\mathrm{\frac{0.0402~bar\dot\ 245~m^3}{\mathrm{8.31~m^3~bar~mol^{-1}~K^{-1}}\dot\ 306~K}=387~mol~\ce{H2O}}}$  

$\mathrm{387~mol~\ce{H2O}\dot\ 18.0~g~mol^{-1}=6.97~kg~\ce{H2O}}$

Step III) Calculate water mass after cooling:  

$T = \mathrm{25^oC=298~K}$  

$P_{saturation} = \mathrm{31.7~mbar^{(1)}}$  

$P = 15\%$ of $P_{saturation} = \mathrm{4.76~mbar}$  

${n=\mathrm{\frac{0.00476~bar\dot\ 245~m^3}{\mathrm{8.31~m^3~bar~mol^{-1}~K^{-1}}\dot\ 298~K}=47.9~mol~\ce{H2O}}}$  

$\mathrm{47.9~mol~\ce{H2O}\dot\ 18.0~g~mol^{-1}=0.862~kg~\ce{H2O}}$ 

Step IV) Calculate total water removed from the house:  

$$\mathrm{6.97~kg~\ce{H2O} - 0.862~kg~\ce{H2O} = 6.11~kg~\ce{H2O}}$$

Reference:
1) Lide, David R., ed. (2006). CRC Handbook of Chemistry and Physics (87th ed.). Boca Raton, FL: CRC Press. ISBN 0-8493-0487-3.
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