Answer
$1559lb$
Work Step by Step
From the solution of 50th problem (last one, image attached below), $F_{AB}$ has the highest value among the three forces/tensions in cables. So, if that is safe all others will be safe. So,
$\\
F_{AB}=0.5133F=800lb \\ F=\frac{800}{0.5133} lb\\ F=1558.5lb$