Answer
$B=3.0 \times 10^{-6} \mathrm{~T}
$
Work Step by Step
The magnitude of the magnitude field at a distance 10 nm away
from atom is,
$$
B=\frac{\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)\left(1.5 \times 10^{-23} \mathrm{~J} / \mathrm{T}\right)}{2 \pi\left(10 \times 10^{-9} \mathrm{~m}\right)}\\=3.0 \times 10^{-6} \mathrm{~T} .
$$
