Answer
After the spheres are separated, the smaller sphere will have a charge of $4.8~\mu C$ and the larger sphere will have a charge of $19.2~\mu C$
Work Step by Step
When the two spheres are brought in contact, the charges will rearrange themselves so there is an equal charge density on each sphere.
Let $q_1$ be the final charge on the smaller sphere and let $q_2$ be the final charge on the larger sphere.
$\frac{q_1}{A_1} = \frac{q_2}{A_2}$
$\frac{q_1}{\pi~r_1^2} = \frac{q_2}{\pi~r_2^2}$
$q_1 = \frac{q_2~r_1^2}{r_2^2}$
$q_1 = \frac{q_2~r_1^2}{(2r_1)^2}$
$q_1 = \frac{q_2}{4}$
$q_2 = 4~q_1$
We can find $q_1$:
$q_1+q_2 = 6.0~\mu C+ 18.0~\mu C$
$q_1+4q_1 = 24.0~\mu C$
$q_1 = \frac{24.0~\mu C}{5}$
$q_1 = 4.8~\mu C$
Then: $~q_2 = (4)(4.8~\mu C) = 19.2~\mu C$
After the spheres are separated, the smaller sphere will have a charge of $4.8~\mu C$ and the larger sphere will have a charge of $19.2~\mu C$
