1.  The first step is to find the resistance of the parallel branch, R_PB.  To do that you have to find the product over the sum.  (30 * 20) / (30 + 20) = 12 ohms

2.  Then you add R1, R4 and R_PB to get the total resistance.  20 + 10 +12 = 42 ohms.

3.  The next thing to find is the total current which is the total voltage(which is given) divided by the total resistance.  120 V / 42 ohms = 2.85 A

4.  Use the Power formula to find the total wattage.  2.85 A * 120 V = 342 W

5.  As is true for all series circuits the current is the same through all parts.  So the two resistors that are in series you have the current for.  R1 and R4 have 2.85 A

6.  Now you can solve for the voltage going through R1 and R4.   R1: 2.85 A * 20 ohms = 57 V & R4: 2.8 A * 10 ohms = 28.5 V

7. The voltage going through all parts of a parallel circuit is the same therefore you can solve for the voltage treating the entire parallel branch as a resistor using 2.8 A and 12 ohms.  2.85 A * 12 ohms = 34.2 V

8.  Now that you have the voltage going through the parallel branch you can solve for the current going through each of the resistors.
I_R1:  34.2 V / 20 ohms = 1.71 A
I_R2:  34.2 V / 30 ohms = 1.14 A

9.  All that is left is solving for the wattage in each resistor using the Power formula. 
P_R1: 2.85 A * 57 = 162.45
P_R2: 1.71 A * 34.2 = 58.48
P_R3: 1.14 A * 34.2 = 38.99
P_R4: 2.85 A * 28.5 = 81.23