Remove the load of buck converter
Your formulas are correct, but I'm not sure where you are getting your numbers from. 

Based on the 18V output voltage, the capacitor energy at the time of the break is 129.6mJ and you are adding 0.77mJ to it, so the capacitor voltage will change from 18V to 18.054V if all the energy of a pulse finds its way into the capacitor. Of course if more pulses come along that will add up quickly at 100kHz. 

In general, \$\Delta V = -V +\sqrt{V^2 + I^2(\frac{L}{C})}\$