In principle it's possible. Let's consider a wire of radius $r$, length $l$ and density $\rho$. If the wire runs magnetic East-West, carries a current $I$ and the horizontal component of the Earth's magnetic flux density is $B_H$, then the vertical component of the Lorentz force on the wire is $B_H I l$. So for levitation,
$$B_H I l=mg=\pi r^2 l\rho g$$
So the current needed would be greater than $$I=\frac{\pi r^2 \rho g}{B_H}$$
In the UK, $B_H=19\ \mu \text T$ and $g=9.8\ \text{N kg}^{-1}$. Let's choose a copper wire, ($\rho=9000\ \text{kg m}^{-3}$) of diameter 1.0 mm. Putting these figures into our formula gives
$$I=3.6\ \text{kA}.$$
Just think what such a large current would do to the wire in a very small fraction of a second!