Metal extracted from molten cryolite is

Cryolite which is ${\mathrm{Na}}_3{\mathrm{AlF}}_6$

Extraction gives

${\mathrm{Na}}_3{\mathrm{AlF}}_6\to 3{\mathrm{Na}}^{+}+{\mathrm{Al}}^{3+}+6{\mathrm{F}}^{-}$

The reaction at the anode is

$6{\mathrm{F}}^{-}⟶{\mathrm{F}}_2+6{\mathrm{e}}^{-}$

The reaction at the cathode is

${\mathrm{Al}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Al}$

Therefore, Option A is the correct answer.