How do I convert AC current at 110v to Watts?
I suggest you read this answer to a previous question.
For a resistive load   


  \$P = V \cdot I \$  


so the 66W would be correct. No need for the \$\sqrt{2}\$; you want the RMS (Root-Mean-Square) value, not the peak value. However, your PC's power supply isn't a pure resistive load and then 


  \$P = V \cdot I \cdot cos(\phi) \$


where \$\phi\$ is the phase difference between current and voltage. You need to measure both simultaneously to determine the phase, something your kill-a-watt does. Since a cos is maximum 1 your power may be lower than the 66W.