here we observed C1 and C2 are parallel connection
Cp=3+6=9nF
Cp and C3 are series connection
the equivalent capacitance Ceq=9*9/9+9=4.5 nF
step;2
now we find the total charge
the total charge Q=4.5*10^-9*10=4.5*10^-8 C
step;3
now we find the charge on each capacitor
charge on capacitor C3=Q3=4.5*10^-8 C
charge on capacitor C1=>Q1=3*4.5*10^-8/3+6=1.5*10^-8 C
charge on capacitor C2=>Q2=6*4.5*10^-8/3+6=3*10^-8 C
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