What is the electron configuration and oxidation state of vanadium in "VBr"_3?

The anion in ${\text{VBr}}_{3}$ is ${\text{Br}}^{-}$, or bromide, contributing a total of a "3- charge to the neutral compound ${\text{VBr}}_{3}$.
To make this compound neutral, we must cancel out the $3 -$ charge with a $3 +$ charge. Naturally, $3 - 3 = 0$. So, the cation is ${\text{V}}^{3 +}$.
Vanadium is atomic number $23$, so its electron configuration is $\left[A r\right] 3 {d}^{3} 4 {s}^{2}$.
Thus, ${\text{V}}^{3 +}$ has an electron configuration of $\textcolor{b l u e}{\left[A r\right] 3 {d}^{2}}$, having removed two $4 s$ electrons and THEN one $3 d$ electron, due to the $4 s$ being higher in energy than the $3 d$.