Correct Answer - A::B::C
Circuit can be drawn as shown in figure

ltbr. In charging of capacitor `R_3` has no role.
In steady state, potential difference acorss capacitor =potentiasl difference across `R_2=E/2`
Therefore , steady state charge across capacitor
`q_0=(CE/2)`
To find time constant of cirucit we will have to short circuit the battery, then we will find net resistance across capacitor.
`R_(n et)=R/2implies tau_C=CR_(n et)=(CR)/2`
`:.` Charging in the capacitor at time t would be
`q=q_0(1-e^((-t)/tau_C)=(CE)/2)(1-e^(-(2t)/(CR)))`