Cr\(_2\)O\(_7\)\(^{2-}\) + 14H\(^+\) + ne-  → 2Cr\(^{3+}\) + 7H\(_2\)O.

All you need to do is to balance the chemical equation, balancing the ionic/ number of moles of the electrons on both sides

Cr\(_2\)O\(_7\)\(^{2-}\) + 14H\(^+\) + ne-  → 2Cr\(^{3+}\) + 7H\(_2\)O.

  (1 x - 2) +  (14 x +1) + n  =  ( 2 x +3)

 - 2  + 14 + n = +6

 12  + n = + 6

What do we need to the +12 on the left hand side to equate +6 on the right hand side

 +12 - 6 = +6

Therefore, number of electrons is 6. i.e 6e\(^-\)