Relatively hard. Won't you say?

Since the distance between each charge is the same, they must be on the vertices of an equilateral triangle. Therefore, the x-components of the electrostatic force cancel out, and the y-components of the forces from 2 and 3 will be the same (since they have the same charge).

Each y-component force will be $F\sin \alpha$ where $F$ = magnitude of the force. Hence:

${\Sigma} F = 2\sin\alpha\times \frac{kQ_{1}Q_{2}}{r^2} = 2\sin 60^{o} \times \frac{(8.99{\times}10^{9})(11{\times}10^{-6})^{2}}{0.15^{2}} = \boxed{83.7N}$ .

First we need our basic formulas $F=\frac { K\times q\times q }{ d^ 2 }$ .Thus we have the equation lets calculate the force as follows ${ F }_{ 21 }={ F }_{ 31 }\frac { 9*10^ 9\times +11\times 10^{ - }6 }{ .15^ 2 } =48.4N$ . Now let us calculate the x , it is obvius that we are going to get ${ F }_{ total x of F 21 }=force\times cos\alpha =48.4*cos(60)=24.2$ and these the other $f 21$ =-24.2 so they cancel each other.Now we need the y forces acting in two directions first ${ F }_{ total\quad x\quad in\quad 21 }=force\times cos\alpha =48.4*sin(60)=41.9$ and the the other y dirction ${ F }_{ total\quad x\quad in\quad 31 }=force\times cos\alpha =48.4*sin(60)=41.9$ so the sum of those forces gives you $\boxed{83.8}$ . Sorry explanation in picture would be more helpful but i don't know how to display picture