The power limit isn't referring to losses or efficiency. It is referring to the point at which the core saturates. Under load, there is still a point where the core saturates, and so no more power can be transferred using the magnetic field.
This power handling capacity is proportional to the amount of iron in the core, and so the power rating is roughly proportional to its weight.
Lets start with a few basics here
$$V = N \dfrac{d}{dt} \Phi = N \cdot A_e \dfrac{d}{dt}B$$
Here \$ B \$ is flux density and for a typical ferrite running at 20kHz I would personally be designing for a flux density swing of around \$200 \text{ mT}\$
Assuming the input voltage is is 27 volts peak to peak maximum, 50% duty cycle @ 20kHz \$ dt = 25 \mu s\$
We now know everything except \$A_e\$ and \$N\$ so we can rearrange the above equation to find \$ N \cdot A_e\$
$$N \cdot A_e = \dfrac{V \cdot dt}{dB} = \dfrac{27 \cdot 25 \times 10^{-6}}{0.2} = 3.38 \times 10^{-3}$$
Note that this product is using Area in SI units (\$m^2\$) and the data sheet probably gives \$A_e\$ in \$\text{mm}^2\$ so we want \$3380000\text{mm}^2\$
In theory any combination of \$N \cdot A_e\$ will work but we note the more turns we have the more area is needed for the windings and the size of the wire we need is set by the current it needs to carry. A typical figure for wire size is to assume a maximum current density of \$ J= 4.5 \cdot \dfrac{\text{amp}}{\text{mm}^2}\$
Given our current of 7.5 amps this requires a wire diameter of \$ \sqrt{\dfrac{4 \cdot I}{J \cdot \pi}}\approx 1.5\text{mm}\$
So we can have a core with a lot of turns and small \$A_e\$ or few turns and big \$A_e\$ the most economic design will be the smallest core into which the turns will fit and if we assume that the wire packs poorly the area required for a single turn is \$A_t = (1.5 \text{mm})^2 = 2.25 \text{mm}^2\$
We can thus work out the area required for the winding \$ A_w = N \cdot A_t \Rightarrow N = \dfrac{A_w}{A_t}\$
Substituting this into the equation above
$$N \cdot A_e = \dfrac{A_w \cdot A_e}{A_t}=\dfrac{V \cdot dt}{dB} \Rightarrow A_w \cdot A_e = \dfrac{A_t\cdot V \cdot dt}{dB} = 7605000 \text{mm}^4$$
Where \$A_e\$ is the core effective area and \$A_w\$ is the available winding area taking into account the bobbin.
Now the coil you have designed will have much more inductance than you want so you will need to introduce an air gap. I usually don't try to calculate this but ask my winding house to gap the core, in the center leg to give me the inductance I want.
The boost converter can be designed in a similar way but I'd ask further advice as at such high frequencies skin and proximity effects are significant.
Best Answer
If you're used to iron cores at mains frequencies, where saturation is the practical limit, or air-cores at RF frequencies, where neither saturation nor core heating occur, then ferrite cores will come as a bit of a surprise.
You have to run a ferrite core at an appropriate B field. There are two limits, one is saturation, which is an absolute 'never exceed' limit. The other is thermal, which is a mid or long term limit, and can be exceeded on any cycle, or for 10s or even 1000s of cycles, depending on the core thermal time constant. If you're running the core at low duty cycle, you could run at well beyond the continuous thermal limit in bursts, as long as you allowed enough time in between bursts so that the average power, leading to core temperature rise, was within limits.
In practice, unless you are running a typical ferrite core at very low frequency, in the 25 kHz region, then core heating will be the limiting factor rather than saturation.
Cycling the core from a positive flux to a negative, and back to positive, dumps a certain amount of thermal energy in the core, from the hysteresis curve. Core heating power is therefore proportional to frequency. While the saturation limit allows you to run the core at 200 mT peak at any frequency, the core heating for continuous use will be acceptable at 25 kHz, and not at 500 kHz, at which frequency you'd either have to reduce the duty cycle, or the flux.