Finding the current dissipated on each resitor
The only voltage source in the system is the 6V battery, so use \$\frac{V}{R_1+R_2+R_3}\$ to find the current \$I\$ going through each resistor. 

Now \$V=IR\$ and \$P = IV\$. You can substitute to get \$P = I^2 × R\$. You can now find the power dissipated by each resistor.