In the tetrafluoroborate anion ($$\left[\mathrm{BF}_{4}\right]^{-}$$), boron is bonded to four fluorine atoms through covalent bonds. Therefore, the covalency of boron in this ion is 4.

The oxidation state of boron can be calculated by considering the charges on the atoms involved in the anion. Fluorine has an oxidation state of -1, and there are four fluorine atoms in the anion, which contributes a total of -4. Since the overall charge on the tetrafluoroborate anion is -1, the oxidation state of boron must be +3 in order to balance the charges.

So, the covalency and oxidation state of boron in $$\left[\mathrm{BF}_{4}\right]^{-}$$ are 4 and 3, respectively.