Writing the balanced oxidation and reduction half-reactions, making the number of electrons same in them and adding both half-reactions gives us the final balanced equation:

$[{\mathrm{Fe}}^{2+}+{\mathrm{C}}_2{\mathrm{O}}_4^{2-}\to {\mathrm{Fe}}^{3+}+2{\mathrm{CO}}_2+3{\mathrm{e}}^{-}]\times 2 \left(\mathrm{oxidation}\right)$

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O} \left(\mathrm{reduction}\right)$

On adding both the reactions, we get,

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+2{\mathrm{Fe}}_{}^{2+}+2{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+14{\mathrm{H}}_{}^{+}\stackrel{}{\to }2{\mathrm{Cr}}^{3+}+2{\mathrm{Fe}}^{3+}+4{\mathrm{CO}}_2+7{\mathrm{H}}_2\mathrm{O}$

Thus, $6$ electrons are donated from oxidized substances to the reduced substance.