Answer
$I = 2400~A$
Work Step by Step
We can find the required current:
$B = \frac{\mu_0~N~I}{L}$
$I = \frac{B~L}{\mu_0~N}$
$I = \frac{B~L}{\mu_0~(L/d)}$
$I = \frac{B~d}{\mu_0}$
$I = \frac{(1.5~T)(2.0\times 10^{-3}~m)}{4\pi \times 10^{-7}~T~m/A}$
$I = 2400~A$
