Cr2O72- + 14H+ + xe- → 2Cr3+ + 7H2O

Consider the total ionic substances on the LHS and that on the RHS

LHS= -2 (from Cr2O7 2-) + 14 (from 14H+) +x  = -2 +14 + x

RHS= 2 x +3 (from 2Cr3+) = +6

Equate LHS and RHS

-2 + 14 + x = +6  (ensure both values on the LHS and RHS are equal)

 +12 + x = +6

 x = 6 -12 = -6 (minus because its an electron)

 thus, x = 6.  Option B is the correct answer (See also image attached)