Response:
a) The diameter available is 0.0384 nm
b)This space is less than the size of a carbon atom, which has a radius of 0.077 nm, indicating that the carbon atom won't occupy these sites.
Clarification:
For BCC iron
According to the information in Appendix B, the lattice parameter (a) is determined to be 0.2866 nm
BCC iron encompasses 4 atomic radii, thus the body diagonal length = 
which represents the atomic radius of BCC iron
4r = 
Substituting the value of (a) from Appendix B, set as 0.2866 nm
4r = 
leading to r = 0.4964 nm / 4 = 0.1241 nm
Refer back to Appendix C, where the atomic radius of BCC iron is stated as 0.1241 nm, assuming the atomic sizes for iron remain consistent.
Thus, the radius ratio = 0.62
According to Figure 3.2, the space necessary for an interstitial at the BCC site exists between atoms located at the FCC site, containing two atoms, each equal to a radius of 0.2482 nm
The diameter of the minimum space available

= 0.2482 nm
With a = 0.2666 nm
therefore
d
= 0.2866 nm - 0.2482 nm = 0.0384 nm
When comparing with the diameter of a carbon atom
This space is smaller than that of a carbon atom which has a radius of 0.077 nm, confirming that the carbon atom will not be able to occupy these positions.