Immediately after the switch is closed, the capacitor has no voltage across it, and can be regarded as a short circuit.
R2 in parallel with R3 is 6*3/(6+3) ie 2 ohms.
The voltage across R2 is 42*2/(2+8) = 8.4
The current in R2 is 8.4/3 amps = 2.8 amps (B)
The current in R3 is 8.4/6 amps = 1.4 amps (C)
The current in R1 is the sum of those, ie, 4.2 amps (A)
When the capacitor is fully charged, it draws no current, and no volts are dropped across R3.
The voltage across the capacitor is 42*6/(6+8) = 18 volts
Charge is c*v = 72 uCoulombs (D)