Me with my physics i do not understand! provide work you.

Fe=K Q1/Q2/d2
Q1 is the first charge
Q2 is the second charge
d is the distance
K= 9x10^9 NM^2/C2
Now let’s plug the numbers
Fe=9x10^9NM^2/C2 (2x10^-4C)(8x10^-4C) / (0.3m^2) you notice we took away the negative charges when we plugged the charges
Ok now we notice that we have C2 which is C to the power 2 we can write it as C^2 and we have two CSU’s beside each one of the charges we can get rid of them all by curtailment
And we can curtailment the M^2and the other M^2
Now we left with only 9x10^9N (2x10^-4)(8x10^-4)/ 0.3
Let’s multiply the (9)(2)(8)=144
And add the exponents (9)+(-4)+(-4)=1
So now we got 144x10N divide by the distance which is 0.3
144x10N / 0.3 = 4800N
Hope it helps u understand :)

Ohm's law tells you that i = v/r where v is equivalent voltage and r is equivalent resistance of the circuit
since all circuits have the same voltage, your question is really asking which circuit has the lowest total resistance, since the lower the resistance, the higher the current
circuit a has a resistance of 5 ohms; circuit c is in series so has a resistance of 10 ohms; circuit b is in parallel and therefore has a resistance of 2.5 ohms (review your notes on adding resistances in parallel and series); therefore b has the highest current