The quantity of electricity in Faraday needed to reduce 1   mol of Cr 2 O 7 2 - to Cr 3 + is

1 faraday of charge is the magnitude of charge of one mole of electrons.

According to the given reaction: 
${\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7^{2-}+\mathrm{ }{6e}^{-}⟶{2\mathrm{C}\mathrm{r}}^{3+}$

Reduction of $1 \mathrm{mol}$ of ${\mathrm{C}\mathrm{r}}_2{\mathrm{O}}_7^{2-}$ to ${\mathrm{C}\mathrm{r}}^{3+}$ requires $6$ moles of electrons. Hence, charge required $=6 \mathrm{Faraday}$