Answer:
185 A
Step-by-step explanation:
In order for the rod to float above the ground, the magnetic force between the two rods must be equal to the weight of the floating rod.
So we can write:

where the term on the left is the magnetic force and the term on the right is the weight, and where:
is the vacuum permeability
is the current in the two rods (they carry the same current)
is the length of the floating rod
r is the distance between the rods
is the mass of the floating rod
g is the acceleration due to gravity
Here we have:




Therefore, solving for I, we find:
