In principle it's possible, provided that the wire is held by some sort of flexible connections so that it wasn't rigidly attached to the rest of the circuit. [If it were so attached, R.W.Bird's answer would apply.]
But let us 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, $$B_H I l>mg\ \ \ \ \ \text{that is}\ \ \ \ \ B_H I l>\pi r^2 l\rho g$$ 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 huge current would do to the wire in a very small fraction of a second!