Answer
The tension in the string is $~~2.4~N$
Work Step by Step
Let $a$ be the acceleration of the system.
We can consider the forces on block A:
$F_A+F_T = m_A~a$
$a = \frac{F_A+F_T}{m_A}$
We can consider the forces on block B:
$F_B-F_T = m_B~a$
$a = \frac{F_B-F_T}{m_B}$
We can equate the two expressions to find the tension in the string $F_T$:
$\frac{F_A+F_T}{m_A} = \frac{F_B-F_T}{m_B}$
$(F_A+F_T)(m_B) = (F_B-F_T)(m_A)$
$(F_T)(m_A+m_B) = (F_B)(m_A)-(F_A)(m_B)$
$F_T = \frac{(F_B)(m_A)-(F_A)(m_B)}{m_A+m_B}$
$F_T = \frac{(24~N)(4.0~kg)-(12~N)(6.0~kg)}{4.0~kg+6.0~kg}$
$F_T = 2.4~N$
The tension in the string is $~~2.4~N$
