Lowest Oxidation number of an atom in a compound A _ 2 B is - 2 ....

${A}_{2}B \to 2{A}^{+} + {B}^{- 2}, {B}^{- 2}$ has complete octet in its dianionic form, thus in its atomic state it has $6$ electrons in its valence shell. As it has negative charge, it has acquired two electrons to complete its octet. If the lowest oxidation state of an atom is $- 2$, this typically implies that the atom can gain two electrons to reach a stable configuration. Therefore, the number of valence electrons in that atom would be six, because gaining two electrons would fill its outer shell.