The covalency and oxidation state respectively of boron in BF 4 - , are

Number of covalent bond formed by Boron is $4$.

${\left[{\mathrm{BF}}_4\right]}^{-}$ Covalency $=4$

The oxidation state of an element represents the charge it would have if all the bonding electrons were assigned to the more electronegative atom in the bond.

Oxidation number of fluorine is -1

Then,

$$\begin{gathered} \mathrm{B} + 4 \mathrm{x} (-1) = -1 \\ \mathrm{B}-4=-1 \\ \mathrm{B}=+3 \end{gathered}$$

Oxidation state $=+3$ for Boron