As Q1 is negative and F12 is attractive, that means Q2 is positive; as Q1 is negative and F13 repulsive, that means Q3 is negative;
Now, use Coulomb’s law:
F12=kQ1Q2/0.01 =5.4 N;
F13=kQ1Q3/0.01 =15 N;
F23=kQ2Q3/0.01 =9 N;
Dividing gives (F13/ F12)= Q3/Q2=15/5.4≈2.78 , and then Q3=2.78Q2
Now, substitute Q3=2.77Q2 into F23 kQ222.77/0.01 =9 , and then Q2≈1.89µC;
and Q3=-2.77Q2=-1.89*2.78 µC=-5.28 µC