ok , so i know the first step is to find t (tau)=RC but should i multiply C with the 6 ohms or with the 3 ohms ? and after that what is the next step ? the voltage signs are on the same place as the capacitor 2mf . plus up negative down

 

There are two time constants, one is before switch is closed that is 6 ohms x 2 mF, after switch is closed, the 3 ohms will be in parallel with 6 ohms... new time constant will be computed...

 

Easy: before the switch is closed, the capacitor is charged to 6ohm*9a=54v.

Right after the switch is closed, that voltage doesn't change. so the instantaneous wattage on the 3ohm resistor is 54^2/3=...

 

If you are unable to determine the time constant by inspection, then it would be good practice to fall back and analyze the circuit by finding the differential equation and seeing what the time constant works out to be. Then see if you can see how that could have been found by inspection.