Rate of a Reaction

For $3A \rightarrow 2B$, the rate of reaction $= -\dfrac{1}{3}\dfrac{d[A]}{dt} = +\dfrac{1}{2}\dfrac{d[B]}{dt}$
$\Rightarrow \dfrac{d[B]}{dt} = -\dfrac{2}{3}\dfrac{d[A]}{dt}$
For every 3 mol of A used, 2 mol of B are formed, so B is formed at $\dfrac{2}{3}$ the rate at which A disappears.

The balanced equation is $5Br^- + BrO_3^- + 6H^+ \rightarrow 3Br_2 + 3H_2O$.
Rate $= -\dfrac{1}{5}\dfrac{d[Br^-]}{dt} = +\dfrac{1}{3}\dfrac{d[Br_2]}{dt}$
$\Rightarrow \dfrac{d[Br^-]}{dt} = -\dfrac{5}{3}\dfrac{d[Br_2]}{dt}$
Note: the source prints the reaction unbalanced (as $H^+ + BrO_3^- + 3Br^- \rightarrow 5Br_2 + H_2O$) and its key gives $\dfrac{d[Br^-]}{dt} = -\dfrac{3}{5}\dfrac{d[Br_2]}{dt}$, which follows from those printed coefficients (3 $Br^-$ and 5 $Br_2$). The equation here is balanced, so the correct answer is changed to $-\dfrac{5}{3}$.