Induction heater design - Current transformer windings ratio
It is not clear where you got the "ESR of the circuit (0.013 Ohms)". Your circuit shows a parallel tank circuit. Perhaps you intended to drive the load as a series circuit?   

Maybe you meant the load resistance (equivalent resistance of the coil). A value of .013 ohms would be reasonable for a coil with 4 uHy.  This gives a coil Q of about 23  (XL/R).    

A coil Q of 23 is quite high for induction heating.   

A coil Q of 23 might be an empty coil or a lightly loaded coil with non-ferrous load. 

Using LTspice :  

Using an input voltage of 18 volts peak keeps your capacitor current (red color) below 50 amps peak.  Green color is current in R1. 

  

The power into the resistor is only about 16 watts. But that is to be expected with such a lightly loaded induction coil.  

To achieve higher power you should try to more tightly couple your heated load to the coil. This will increase R and decrease the coil Q (XL/R).   

You are likely to need capacitors that can handle higher currents if you expect to achieve higher power to your heated load.