Answer
$F=2.46N$
Work Step by Step
The force that link 3 exerts on link 2 can be calculated by adding the weights of link 1 and link 2 to the force required to accelerate the links 1 and 2 together at $2.50m/s^{2}$,
$F_{32}=(m_{1}+m_{2})g + (m_{1}+m_{2})a$
$F_{32}=(m_{1}+m_{2} ) (g +a)$
$F_{32}=[(0.200kg) (9.8m/s^{2}+2.5m/s^{2})]=2.46 N$
