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My solution above was not posted completely due to size limit I guess and only showed heat load through roof which might have confused you more( sorry about that). 


I was trying to post a solution with heat load through roof, walls and windows. Anyways let's just talk about windows:


The CLTD is only given for walls and roof. If it is given for windows then your approach would be right to calculate heat load through windows by using the following equation:


Q=Qrad +Qcond


Q=(S)(SHGF)(A)+UA(CLTD)


but in this question CLTD is not given for windows and just like for walls it will depend on glass material and properties etc. We can't assume CLTD for walls to be same as for windows.


In a situation like this, it is often a practice to use delta T between outdoor and indoor to calculate conduction load through windows.


hope that clears the confusion.


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