Given the cation radius is 148 pm and the anion radius is 195 pm, determine the coordination number and crystal structure.

The radius of the cation A$^+$ ($r_c$) is $148 \, \text{pm}$.
The radius of the anion B$^-$ ($r_a$) is $195 \, \text{pm}$.

The radius ratio rules for ionic compounds are:

• $0.155 - 0.225$: Coordination number 3 (Trigonal planar)
• $0.225 - 0.414$: Coordination number 4 (Tetrahedral)
• $0.414 - 0.732$: Coordination number 6 (Octahedral, NaCl type)
• $0.732 - 1.000$: Coordination number 8 (Cubic, CsCl type)

The calculated radius ratio of approximately 0.759 falls within the range $0.732 - 1.000$. This indicates a coordination number of 8.

A coordination number of 8 for an ionic compound typically corresponds to a cubic arrangement, such as the CsCl structure. In the CsCl structure, one type of ion (e.g., cation) is located at the body center of a cube formed by eight ions of the other type (e.g., anions). This arrangement is often referred to as a Body-Centered Cubic (BCC) structure in the context of ionic compounds.

Comparing this with the options:

• (a) Body centred cubic: This matches the expected structure for a coordination number of 8.
• (b) Face centred cubic: This corresponds to a coordination number of 6 (e.g., NaCl type).
• (c) Tetrahedral: This corresponds to a coordination number of 4.
• (d) Octahedral: This corresponds to a coordination number of 6.

Therefore, the ionic compound AB will have a Body-Centered Cubic structure.

The final answer is $\boxed{\text{a) Body centred cubic}}$.