Answer
$F\left( 6\right) > F\left( 5\right) > F\left( 4\right) > F\left( 3\right) > F\left( 2\right) > F\left( 1\right) =0$
Work Step by Step
Net torque is zero at point $A$ and $B$. If we want to calculate force on point $B$, we need to take point $A$ as the reference point.
So if the safe creates a bigger torque on point $A$ then it creates a greater force on point $B$ because this torque will be equal in magnitude to the torque that point $B$ applies on point $A$ and the distance from point B to point A doesnt change.
Torque created by the safe on point A depends on the distance of the safe from the point.
So, if we want to compare forces, we basically need to compare distances of safe to point A and we get,
$F\left( 6\right) > F\left( 5\right) > F\left( 4\right) > F\left( 3\right) > F\left( 2\right) > F\left( 1\right) =0$
