What is the power dissipated by R2, R4, and R6?

ANS:A - P2 = 417 mW, P4 = 193 mW, P6 = 166 mW

Simple solution:-

First find the total resistance of the circuit i.e., 6.25 kiloohm (hope you can find that).
Then find total current i.e., 111/6.25 = 17.73mA.

This current will only flow through 3kiloohm resistor as it is in series with the voltage source.
So the voltage drop across 3kiloohm resistor will be =current through resistor*resistance
=17.73*3 = 53.29 v.

So the voltage at first junction after 3kiloohm resistor is :- 111-53.29=57.79v.
So the voltage across 8kiloohm resistor will be 57.79-0=57.79 v.

We subtract it with zero because the other end of 8kiloohm resistor is at 0 vas it is connected to -ve end of voltage source.

Now power through 8kiloohm resistor is = V(8kiloohm)*V(8kiloohm)/resistance.
=57.79*57.79/8 =417 mW.

SO OPTION A is correct.