smc

Since $\sin$ can be seen as "opposite over hypotenuse" in a right triangle, we can label the diagram as shown. By the Pythagorean Theorem, we have: $$y^2+(2x{)}^2=(3x{)}^2$$ $$y=\sqrt{9x^2-4x^2}$$ $$y=\sqrt{5x^2}$$ $$y=x\sqrt{5}$$ so $\tan B=\frac{x\sqrt{5}}{2x}=\frac{\sqrt{5}}{2}$, which is choice $\boxed{\frac{\sqrt{5}}{2}}$.