Answer
$B = 0.13~T$ (out of the page)
Work Step by Step
In order to levitate the wire, the upward force from the interaction of the current with the magnetic field must be equal in magnitude to the weight of the wire.
We can find the magnetic field strength:
$F_B = F_g$
$I~L~B = mg$
$B = \frac{mg}{I~L}$
$B = \frac{(0.0020~kg)(9.80~m/s^2)}{(1.5~A)(0.10~m)}$
$B = 0.13~T$
By the right hand rule, the magnetic field must be directed out of the page.
