Your approach is naive. The following may also be subject to errors.

For strictly discontinuous operation with a single or two switch flyback you assume 50% duty cycle at minimum input voltage to allow for reset. The minimum input voltage is set by design of the input filter and smoothing. Let's say you have 110VAC and use a voltage doubler. Peak is 311V but you design for say 250V. It depends on how you size the filter capacitors for ripple and if required a half cycle line drop out.

Pick your required output power. 100W. Assume an efficiency of 80% and your input power becomes 120W. The average input current from 250V will be 120/260 or 0.46 amps. The input current waveform is a 50% duty cycle triangle wave so it's peak is 4 times the average or 1.85A. Design for a 20% overload and you set the current limit in your supply at 2.3A. This is also the peak primary current. The RMS primary current will be 0.41 times the peak for 50% duty cycle or 0.76A

If your operating frequency is 100KHz 50% duty cycle will be a 5uS on time. During this time your primary is set through 250V. The primary inductance will be given by...

Lpri = Vin.Ton/dI = (250 x 5E-6)/2.3

Lpri = 540uH

You assign equal, half the available, winding area to the primary and secondary(ies). You estimate the required core area product from...

Aw.Ae = Lpri.Ipk.Irms/Bpk.J.Kw.Kcu.Ki

Aw is the bobbin winding area.

Ae is the core effective area.

Lpri is your target primary inductance.

Ipk is the peak primary current.

Irms is the RMS primary current.

Bpk is the peak/suturation flux density of the core.

J is the operating current density of the winding.

Kw is the primary utilisation of the winding area.

Kcu is the wire utilisation of the winding area.

Ki is for margin isolation.

Bpk may be loss limited but assume 300mT

Kw is 0.5 Half assigned to primary.

Kcu is 0.7 for round wire.

Ki is a guess of 0.8. 20% loss for two times 3mm margins 6mm total.

J is 2E6 assuming a 40C rise and convection cooling.

Some of these numbers are subject to wide variations depending on the application. In particular the number chosen for J. For a simple inductor operating with 20% ripple current 4E6 would be a reasonable value. In this case winding currents are discontinuous and the AC losses go up as a result of penetration and layer effects so I have used 2E6.

Aw.Ae = (540E-6 x 2.3 x 0.76)/(300mT x 2E6 x 0.5 x 0.7 x 0.8)

Aw.Ae = 562E-9

Assuming Aw = Ae results in Ae = 75mm^2

The minimum number of primary turns you need is given by...

Npmin = Lpri.Ipk/Bpk.Ae

Npmin = 540E-6 X 2.3A/300mT x 75E-6

Npmin = 55

https://ferrite.ru/uploads/pdf/products/etd/etd_29_16_10.pdf
Looks fairly close and is available gapped to 200nH per root turn. 50 turns would give an Lpri 500uH which apparently would violate my original sums but these things are variable. It's all an approximation but the idea of the approximation is to quickly home in on a possible solution. Once there or close you iterate around that solution to check as to its viability.

Unfortunately that includes consideration of other things and, for the moment, I am not going to do that. I would tentatively be happy to put 100W through an ETD29 core in an offline application using a discontinuous flyback topology.