Answer
$F = 4.682 \times 10^{-19} N$
Work Step by Step
Sphere $A : Q$
Sphere $ B : -\frac{Q}{4}$
Sphere $C : \frac{Q}{2}$
When sphere C touches sphere A, they divide up their total charge
$\frac{Q}{2} + Q = \frac{3Q}{2}$
Divide this equally by them both
$ \frac{3Q}{2} \div 2 = \frac{3Q}{4}$
The magnitude of the force of attraction between A and B becomes
$F = k[\frac{(3Q/4)|Q/4|}{d^2} ]$
$F = k[\frac{(3Q^2/16)}{d^2} ]$ where $ d = 1.20 m$ and $Q = 2.00 \times 10^{-14} C$
$F = (8.99 \times 10^9 N . m^2 / C^2)[\frac{(3/16)(2.00 \times 10^{-14} C)^2}{(1.20m)^2} ]$
$F = 4.682 \times 10^{-19} N$
