Answer
Magnitude: $\frac{2kq^2}{r^2}$
Direction: Upwards
Work Step by Step
All the charges are the same on opposite ends of the circle so they cancel out except for 2q charge located at the top of the inner circle. Therefore the $F_{net}$ would be $\frac{k\times q\times2q}{r^2} = \frac{2kq^2}{r^2}$
Also, the force would be upwards in the direction of 2q.
