Answer
The tension in the cord at the right is $~~81.7~N$
Work Step by Step
Since $m_C \gt m_A$, then block C will accelerate downward, block B will accelerate to the right, and block A will accelerate upward.
We can find the magnitude of the acceleration:
$m_C~g - m_A~g = (m_A+m_B+m_C)~a$
$a = \frac{(m_C - m_A)~g}{m_A+m_B+m_C}$
$a = \frac{(10.0~kg - 6.00~kg)~(9.8~m/s^2)}{6.00~kg+8.00~kg+10.0~kg}$
$a = 1.63~m/s^2$
We can consider the forces on block C to find the tension $F_T$ in the cord at the right:
$m_C~g-F_T = m_C~a$
$F_T = m_C~(g-a)$
$F_T = (10.0~kg)~(9.8~m/s^2-1.63~m/s^2)$
$F_T = 81.7~N$
The tension in the cord at the right is $~~81.7~N$