I've got an answer of 24 turns but this doesn't feel right to me.
Could someone double check my calculations?
The data sheet for the toroid tells us that \$A_L\$ is this: -
And the relationship between \$A_L\$ and inductance (L) is this: -
$$L = N^2\cdot A_L$$
So, if you have 24 turns and \$A_L\$ is 1170 \$nH/turn^2\$, inductance is 673.92 uH.
Don't derive this yourself because the data sheet has the numbers directly "inside" \$A_L\$.
If you need 150 uH then you'll need about 11 turns (141.6 uH)
If you are operating closer to saturation (as you appear to be saying) then \$A_L\$ reduces proportionally with permeability. So, with a H-field of 60 At/m, the relative permeability has dropped from 2300 to 1000 hence, \$A_L\$ drops by the same amount from 1170 to 507.
Now if 24 turns are used, I calculate an inductance of 293 uH (still higher than your calculations). I estimate 17 turns would be needed to give 146.5 uH.
However, the bigger question for me is why you are using this toroid to make a boost converter. At a 60 Ah/m H-field and an effective core length of 30.1 mm (see above), the MMF (magneto motive force) becomes 60 x 0.0301 = 1.806 ampere-turns AND, given that you have possible 17 turns, the maximum current you can push through the toroid (for reasonable efficient operation) is 106 mA and this is a pretty low power boost converter in my humble opinion.