Answer
The ratio of the magnitude of the force on either sphere after they were touched to that before they were touched is $\frac{4}{5}$
Work Step by Step
We can find the original magnitude of the electric force:
$F_1 = \frac{k~\vert q_1 \vert ~\vert q_2 \vert}{L^2}$
$F_1 = \frac{k~(5.0~\mu C)(1.0~\mu C)}{L^2}$
After the two spheres touch, the charge on each sphere is $+2.0~\mu C$
We can find the new magnitude of the electric force:
$F_2 = \frac{k~\vert q_1 \vert ~\vert q_2 \vert}{L^2}$
$F_2 = \frac{k~(2.0~\mu C)(2.0~\mu C)}{L^2}$
We can find the ratio $\frac{F_2}{F_1}$:
$\frac{F_2}{F_1} = \frac{\frac{k~(2.0~\mu C)(2.0~\mu C)}{L^2}}{\frac{k~(5.0~\mu C)(1.0~\mu C)}{L^2}}$
$\frac{F_2}{F_1} = \frac{(2.0~\mu C)(2.0~\mu C)}{(5.0~\mu C)(1.0~\mu C)}$
$\frac{F_2}{F_1} = \frac{4}{5}$
The ratio of the magnitude of the force on either sphere after they were touched to that before they were touched is $\frac{4}{5}$
