Answer
$\bf 2.4\;\rm kA$
Work Step by Step
We know that the magnetic field created by a solenoid is given by
$$B=\dfrac{\mu_0NI}{l}$$
And we need to find the current needed, so
$$I=\dfrac{lB}{\mu_0 N}$$
where $N$ is the number of turns which is given by the total length of the wire divided by the diameter of the wire.
$$I=\dfrac{lB}{\mu_0 (l/D)}$$
$$I=\dfrac{ DB}{\mu_0 }$$
Plug the known;
$$I=\dfrac{ DB}{\mu_0 }$$
$$I=\dfrac{ (2\times10^{-3})(1.5)}{(4\pi \times10^{-7})}$$
$$I=\color{red}{\bf 2.39}\;\rm kA$$
