Let $\angle B$-bisector intersect $AC$ at $D$ and Let bisector $BD$ intersect the circle at $E$ By angle bisector theorem, $\frac{AB}{BC}=\frac{AD}{DC} \implies \frac{AD}{DC}=\frac{1}{3} \implies \boxed{3AD=DC}$ For the side $AC$, we have our cosine rule, $AC=\sqrt{3+27-9}=\boxed{\sqrt{21}}$ Hence, $AD=\frac{\sqrt{21}}{4} \text{ and } DC=\frac{3\sqrt{21}}{4} $ For the $BD$, $BD=\frac{2\cdot AB \cdot BC}{AB+BC} \cos{\frac{B}{2}}= \boxed{\frac{9}{4}}$ By power-point theorem, $BD \cdot DE=AD \cdot DC\implies DE=\frac{4}{9} \cdot \frac{\sqrt{21}}{4} \cdot \frac{3\sqrt{21}}{4} =\boxed{\frac{7}{4}}$ Hence, the chord is $BE=BD+DE=\frac{9}{4}+\frac{7}{4}=\boxed{4}$