Answer
The tension in the rope at its midpoint is $~~\frac{(F)~(2M+m)}{2~(M+m)}$
Work Step by Step
We can find the acceleration of the rope and block:
$F = (M+m)~a$
$a = \frac{F}{M+m}$
Note that the tension $F_T$ in the middle of the rope provides the force to accelerate the left half of the rope and the block toward the right. we can find $F_T$:
$F_T = (M+\frac{m}{2})~a$
$F_T = (M+\frac{m}{2})~(\frac{F}{M+m})$
$F_T = (\frac{2M+m}{2})~(\frac{F}{M+m})$
$F_T = \frac{(F)~(2M+m)}{2~(M+m)}$
The tension in the rope at its midpoint is $~~\frac{(F)~(2M+m)}{2~(M+m)}$
