Imagine adding electrons to the pin until the negative charge has the very large value 3.00 mc. how many electrons are added for every 109 electrons already present?

The formula for this problem would be this:

electrons added = Q / e = 3 x 10^-3 C / (1.6 x 10^ - 19 C / electron)

number of electrons added = 1.875 x 10^16

1.875 x 10^16 / 2.62 x 10^24 = 7.156 x10^-9

so 7.156 electrons for every 10^9 are already present.