What is the magnitude of the electric force on charge A in (Figure 1) ?

Note that by Coulomb's Law,

F = kq1q2/r^2

where

k = 8.99E9 Nm^2/kg^2
q1, q2 = the two interacting charges
r = the distance of separation of the two charges

Thus, for force of B on A points to the right, as opposite charges attract.
qA = 1.0E-9 C
qB = -1.0E-9 C
r = 0.01 m

Thus,

F(B on A) = 8.99E-5 N

Now, for force of C on A points to the left, as like charges repel.

Thus, as

qA = 1.0E-9 C
qC = 4.0E-9 C
r = 0.02 m

F(C on A) = -8.99E-5 N [the negative sign because it is to the left]

Thus,

Fnet (A) = F(B on A) + F(C on A)

Fnet (A) = 0.00 N [ANSWER]