Answer
The tension in cord C is 34 N
Work Step by Step
We know that the magnitude of the acceleration will be the same for both blocks. Let $F_T$ be the tension in the cord connected to both blocks.
For the 1.2-kg block:
$F_T - mg = ma$
$a = \frac{F_T-mg}{m}$
For the 3.2-kg block:
$Mg - F_T = Ma$
$Mg - F_T = M(\frac{F_T-mg}{m})$
$Mmg - mF_T = MF_T - Mmg$
$F_T = \frac{2Mmg}{M+m}$
$F_T = \frac{(2)(3.2 ~kg)(1.2 ~kg)(9.8 ~m/s^2)}{3.2 ~kg+1.2 ~kg}$
$F_T = 17 ~N$
The tension in the cord connected to both blocks is 17 N.
So, the tension in cord C is $2\times F_T$ which is equal to 34 N
