The charge required to deposit one mole aluminum from the electrolysis of  $A{l}_2{O}_3$

From $A{l}_2{O}_3\underset{ }{\overset{ }{\to }}2Al$  to obtain one mole of  $Al$
We consider 
$2A{l}^{+3}+6e^{-}\underset{ }{\overset{ }{\to }}2Al$
So,  $6F$ of charge is required to obtain 2 moles of  $Al$ from $A{l}_2{O}_3$ . 
Thus $3F$  of charge is required to obtain 1 mole of  $Al$  from  $A{l}_2{O}_3$.