transformer magnetizing inductance vs magnetizing current
If the primary inductance is 2.86 mH and you apply 400 volts for 6 us (50% of the period of 83 kHz) then the magnetization current will rise to 400 x 6 us/ 2860 uH = 0.839 amps over that period. 

The above calculation is a re-arrangement of V = L di/dt.

Given that you have 26 turns, that's a peak ampere-turns (aka magneto motive force) of 21.8. Given that the effective length of an ETD59 core-set is 139 mm, then the H-field will be 21.8/0.139 = 157 ampere-turns per metre.

If you are using N27 material, the peak flux will be close to saturation: -



I've drawn the red line roughly indicating where the H field is 157 At/m

If you doubled the turns, the inductance would be 4 times greater and the current would be a quarter of what it previously was after 6 us. However, turns have doubled so, in effect, there is only a net reduction in H-field of 2 but this would be better positioned to give significantly less saturation.

See also THIS question and answer for a similar situation to yours.