$\therefore$ Applying Ampere's law.

$\oint \overrightarrow{\mathrm{B}} \cdot \mathrm{d} \vec{\ell}=\mu_{0} \mathrm{i} \Rightarrow \mathrm{B}_{\mathrm{A}} 2 \pi \frac{\mathrm{a}}{3}=\mu_{0} \mathrm{J} \pi\left(\frac{\mathrm{a}}{3}\right)^{2}$

$\therefore B_{A}=\frac{\mu_{0} J a}{6}$

Similarly, $B_{B}=\frac{\mu_{0} J a}{4}$

$\frac{B_{A}}{B_{B}}=\frac{\mu_{0} J a \times 4}{\mu_{0} J 6 a}=\frac{2}{3}$