The core losses agree for both datasheets: about 5.2W for 100mT, 100kHz. But the core datasheet says nothing about the material at 200mT, 300kHz, so there you need to use the material information. First check both datasheets agree for a known datapoint, then extrapolate to your working conditions.
In
this third datasheet you have the formulae for inductance and magnetizing current. Let us work at 25ºC, and define k1= 508.0, k2=-0.708, k3=812.0, k4=-0.796.
For a coil of N=100 turns:
The \$A_L\$ value for a gap \$s\$ is: \$\displaystyle A_L\ = \ k_1\,s^{k2}\$
And the DC current \$I_{DC}\$ at which \$A_{L}\$ drops by 10%, marking the onset of saturation, is given by: \$I_{DC} \ = \ \displaystyle\left(\frac{0.9\cdot A_{L}}{k_3}\right)^{1/k_4}\$
So, let us assume you have a gap of s=1.5mm. Then, AL = (1.5**-0.708)*508.0 = 381, which exactly agrees with the datasheet.
Then, the magnetizing current for saturation is: Idc = (0.9*381.0/812.0)**(1.0/-0.796) = 2.95 A, near 3A.
Now, these formulas are for N=100 turns. If you want 3.45uH in each winding, since \$L = A_{L}\cdot N^2\$, then \$N =\displaystyle \sqrt{L/A_{L}}\$ (in nanohenries!), so N = sqrt(3450.0/381) = 3, ok.
I think that in your circuit the 3+3=6 turns work as a single inductor, so the DC current for 10% saturation at 25ºC should be 2.95*100/6 = 49 A. I think you are safe enough at 5A max.
Be warned, though, that I'm neither an EE, but a mathematician.