The correct options are
A The potential at junctions A,B are 8 V and 2 V respectively
D The voltage across capacitance is 6 V and charge stored in it is 6 μC
After a long time, capacitor is fully charged and no charge flows through it.
VA→ potential at left junction
VB→ potential at right junction
Current through left branch,
IA=10 V5 Ω=2 A
⇒VA=10−2(1)=8 V
Current through right branch
IB=10 V10 Ω=1 A
⇒VB=10−8(1)=2 V
⇒VA−VB=(8−2) V=6 V
Charge stored in capacitor, Q=C(VA−VB)=6 μC