a
Let current density be J.

\(\therefore\) 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, \(B_{B}=\frac{\mu_{0} J a}{4}\)

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