To find the electric potential at A you would need to add up the electric potentials of the two point charges affecting A, namely q1 and q2.

The formula for electric potential of point charge is

V = (ke*q)/r

Now you must apply this equation to q1 and q2:

V1 = (ke*q1)/r and V2 = (ke*q2)/r

plug in the values which are:

where ke = 8.988*10^9 N*m^2/C^2 (found in textbook)

q1 = -12*10^-9 C

q2 = 27.5*10^-9 C

r = 4*10^-2 m

(*note i have already converted all units to SI units)

V1 = (8.988*10^9 N*m^2/C^2 * -12*10^-9 C) / 4*10^-2 m = -2696.4 V

V2 = (8.988*10^9 N*m^2/C^2 * 27.5*10^-9 C) / 4*10^-2 m = 6179.26 V

Now add up V1 and V2 to get the total electric potential at point A.

V1 + V2 = (-2696.4 V) + (6179.26 V) = 3482.9 V ~ 3.48 kV

Part B is easy to find now that we have solved part A. Since point B is at a distance half of the original distance from both charges, you would essentially be dividing the answer by (1/2), which is the same is multiplying by 2,

and so the electric potential at B is equal to 2*3.48 kV = 6.96 kV