Answer
$B \approx30.7^\circ$, $C =99.3^\circ$, $c\approx3.86$. Only one triangle
Work Step by Step
Step 1. Based on the given conditions, use the Law of Sines, we have $\frac{sinB}{2}=\frac{sin50^\circ}{3}$, thus $B=sin^{-1}(\frac{2sin50^\circ}{3})\approx30.7^\circ$ or $180-30.7=149.3^\circ$ (discard this one as A+B>180)
Step 2. We have $C=180-50-30.7=99.3^\circ$
Step 3. We have $\frac{c}{sin99.3^\circ}=\frac{3}{sin50^\circ}$, and $c=\frac{3sin99.3^\circ}{sin50^\circ}\approx3.86$. Only one triangle is possible.