Answer
$r_{}=4.80 cm .$
Work Step by Step
We now look for a solution in the exterior region, where the field is inversely proportional to $r$
$$
\frac{B}{B_{\max }}=\frac{3.00 \mathrm{mT}}{12.0 \mathrm{mT}}=\frac{R}{r}
$$
which yields $r=4 R=4(1.20 \mathrm{~cm})=4.80 \mathrm{~cm}$.
