Correct Answer - d
In steady state, there will be no current in the capacitro branch .Net resistacne of the circuit `R=1+1+0.5=2.5omega`
Current drawn from the cell `i=(V)/(R )=(2.5)/(2.5)=1A`
Potential drop across two parallel bracnhes
`V=E-ir=2.5-1xx0.5=2.5-0.5=2.0V`
So, charge on the capacito plates
`q=CV=5xx2=10muC`