Answer
1400 amps, which does not seem feasible.
Work Step by Step
For the wire to float, the upward magnetic force balances the wire’s weight.
$$F_B=mg$$
The wire is perpendicular to the magnetic field, so the maximum magnetic force for the current and field is applicable.
The mass of the wire is the density of copper multiplied by the volume of the wire. Assume a length L of wire.
$$ILB=\rho \pi (\frac{1}{2}d)^2Lg$$
$$I=\frac{\rho \pi d^2Lg}{4B}=\frac{(8900kg/m^3)\pi(0.001m)^2(9.80m/s^2)}{4(5.0\times10^{-5}T)}=1400A$$
Such a large current, 50-100 times larger than a typical current in a household appliance’s electrical circuit, would probably melt such a thin wire.
