Intuitively, it seems the fly should land in the opposite corner.
If we unfold the room along the edge between short wall and roof, the spider travels 8 feet across and 24 feet along, for a distance $\sqrt{24^2+8^2}$ feet
However, we can improve things for the spider by
Unfolding the room along the join between the long wall and the roof gives us a distance of $\sqrt{16^2+16^2}$
Prompted by Johannes' comment that this can be beaten:
As shown below, if the fly moves some distance along the short wall it simultaneously improves the worse solution given and lengthens the improved solution. When they are equal the distance should be maximised.

The distances are
$d_1=\sqrt{16^2+(16+x)^2}$,
$d_2=\sqrt{24^2+(8-x)^2}$
Which are equal to $\sqrt{24^2+(16/3)^2} \approx 24.585$
at x=8/3