Calculating the number of turns for my inductor
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.