The box is in equilibrium when the sum of forces on the box in vertical as well as horizontal direction is zero.
Assuming the force in right as positive and force in left direction as negative.
Assuming the force in up as positive and force in down direction as negative.
For Box 1:
[tex]\sum F_{x}[/tex] = Net force along horizontal direction = 5 - 3 = 2 N
[tex]\sum F_{y}[/tex] = Net force along vertical direction = 5 - 3 = 2 N
Since [tex]\sum F_{x}[/tex] ≠ 0 and [tex]\sum F_{y}[/tex] ≠ 0
hence the box is not in equilibrium
For Box 2:
[tex]\sum F_{x}[/tex] = Net force along horizontal direction = 5 - 5 = 0 N
[tex]\sum F_{y}[/tex] = Net force along vertical direction = 2 - 2 = 0 N
Since [tex]\sum F_{x}[/tex] = 0 and [tex]\sum F_{y}[/tex] = 0
hence the box is in equilibrium
For Box 3:
[tex]\sum F_{x}[/tex] = Net force along horizontal direction = 5 - 5 = 0 N
[tex]\sum F_{y}[/tex] = Net force along vertical direction = 0 - 5 = - 5 N
Since [tex]\sum F_{x}[/tex] = 0 and [tex]\sum F_{y}[/tex] ≠ 0
hence the box is not in equilibrium
For Box 4:
[tex]\sum F_{x}[/tex] = Net force along horizontal direction = 2 - 2 = 0 N
[tex]\sum F_{y}[/tex] = Net force along vertical direction = 2 - 5 = - 3 N
Since [tex]\sum F_{x}[/tex] = 0 and [tex]\sum F_{y}[/tex] ≠ 0
hence the box is not in equilibrium
hence only second box is in equilibrium.
