The situation as described in the problem is shown in above figure.
Let the charges on each point charges (A,B,C) be #qC#
In #Delta OAB,/_OAB=1/2(180-45)=67.5^@#
So #/_CAB=67.5-45=22.5^@#
#/_AOC=90^@#
So #AC^2=OA^2+OC^2=2L^2#
#=>R^2=2L^2#
For #Delta OAB,#
#AB^2=OA^2+OB^2-2OA*OBcos45^@#
#=>r^2=L^2+L^2-2L^2xx1/sqrt2=L^2(2-sqrt2)#
Now forces acting on A
Electrical repulsive force of B on A
#F=k_eq^2/r^2#
Electrical repulsive force of C on A
#F_1=k_eq^2/R^2#
where #k_e="Coulomb's const"=9xx10^9Nm^2C^-2#
#F/F_1=R^2/r^2=sqrt2/(2-sqrt2)=(sqrt2(2+sqrt2))/((2+sqrt2)(2-sqrt2))#
#=(2sqrt2+2)/2=sqrt2+1#
And #T="Tension on string"#
Considering equilibrium of forces acting on A we can write
For vertical forces on A
#Tcos45+Fsin22.5=mg#
#=>T/sqrt2=mg-Fsin22.5........[1]#
For horizontal forces on A
#Tsin45=Fcos22.5+F_1#
#=>T/sqrt2=Fcos22.5+F_1........[2]#
Comparing [1] an [2] we get
#Fcos22.5+F_1=mg-Fsin22.5#
#=>Fcos22.5+Fsin22.5+F_1=mg#
#=>F(cos22.5+sin22.5)+F_1=mg#
#=>F(sqrt(cos^2 22.5+sin^2 22.5+2sin22.5xxcos22.5))+F_1=mg#
#=>F(sqrt(1+sin45))+F_1=mg#
#=>F(sqrt(1+1/sqrt2))+F_1=mg#
#=>F(sqrt((2+sqrt2)/sqrt2))+F_1=mg#
#=>F_1xx(sqrt2+1)(sqrt((2+sqrt2)/sqrt2))+F_1=mg#
#=>F_1[(sqrt2+1)(sqrt((2+sqrt2)/sqrt2))+1]=0.1xx9.81#
#=>F_1xx6.47=0.1xx9.81#
#=>k_eq^2/R^2=(0.1xx9.81)/6.47~~0.152#
#=>q=Rxxsqrt(0.152/k_e)#
#=>q=sqrt2Lxxsqrt(0.152/k_e)#
#=>q=sqrt2xx0.3xxsqrt(0.152/(9xx10^9))C=1.74muC#