Answer:
The charge is  [tex]Q = 2.177 *10^{-9} \ C[/tex]
Explanation:
From the question we are told that
   The electric potential is  [tex]V = 25.0 \ kV = 25.0 *10^{3}\ V[/tex]
   The  mass of the drop is  [tex]m = 0.168 \ m g = 0.168 *10^{-3} \ g = 0.168 *10^{-6}\ kg[/tex]
   The  speed is  [tex]v = 18.8 \ m/s[/tex]
Generally the charge on the paint drop due to the electric potential which will give it the speed stated in the question  is mathematically represented as
    [tex]Q = \frac{m v^2 }{ 2 * V }[/tex]
Substituting values
   [tex]Q = \frac{0.168 *10^{-6} (18)^2 }{ 2 * 25*10^3 }[/tex]
    [tex]Q = 2.177 *10^{-9} \ C[/tex]
