Answer
$1.68\times10^{-4}eV$.
The $m_s=+\frac{1}{2}$ state has lower energy.
Work Step by Step
The energy expressions are considered in example 41.6, at the top of page 1382.
In this case, because the field is in the –z direction, the energy of the $m_s=+\frac{1}{2}$ state is lowered, and the energy of the $m_s=-\frac{1}{2}$ state is raised by the same amount.
The energy difference between them is twice the amount of their individual shifts.
$\Delta U=2(9.285\times10^{-24}J/T)(1.45T)=2.692\times10^{-23}J$
$=1.68\times10^{-4}eV$
As stated above, the $m_s=+\frac{1}{2}$ state has lower energy.
