Answer
$F=3.69N$
Work Step by Step
The force that link 4 exerts on link 3 can be calculated by adding the weights of link 1 through link 3 to the force required to accelerate the links 1 through 3 together at $2.50m/s^{2}$,
$F_{43}=(m_{1}+m_{2}+m_{3})g + (m_{1}+m_{2}+m_{3})a$
$F_{43}=(m_{1}+m_{2}+m_{3} ) (g +a)$
$F_{43}=[(0.300kg) (9.8m/s^{2}+2.5m/s^{2})]=3.69 N$
