Let us take an gaussian surface as shown. This proves the charge distribution in the fig. as E inside the conductor is 0.
Also, E(p)=0
so,x=x1
ΣQ= x+y-y+z-z+b-b+c-c+w-w+x1
x+x1=ΣQ
Hence, x=x1=ΣQ/2
In the question, ΣQ=q
Therefore, charge on the outer surface will be q/2
(Plz let me know if there is any mistake..) :)