Explanation:

Given:
Density of NaCl = 3.165 g/cm^3

To find:
Distance between Na+ and Cl- ions in NaCl

Formula:
Density = (atomic mass of Na + atomic mass of Cl) / (volume of the unit cell * Avogadro's number)

Let's assume that the NaCl crystal structure is face-centered cubic (FCC). In an FCC unit cell, there are 4 Na+ ions and 4 Cl- ions.

1. Finding the volume of the unit cell:
In an FCC unit cell, the volume of the unit cell is given by:
Volume of unit cell = (4 * r^3) / 3
where r is the radius of the Na+ ion (same as the Cl- ion).

2. Finding the mass of the unit cell:
Mass of unit cell = (atomic mass of Na * 4) + (atomic mass of Cl * 4)

3. Finding the volume of the unit cell:
Density = Mass of unit cell / Volume of unit cell
3.165 g/cm^3 = [(atomic mass of Na * 4) + (atomic mass of Cl * 4)] / [(4 * r^3) / 3]

4. Rearranging the equation to solve for r:
r^3 = [(3.165 * 3 * atomic mass of Na) / (4 * atomic mass of Cl)]

5. Taking the cube root of both sides:
r = [(3.165 * 3 * atomic mass of Na) / (4 * atomic mass of Cl)]^(1/3)

6. Finding the distance between Na+ and Cl- ions:
The distance between Na+ and Cl- ions is equal to twice the radius (2r).

7. Calculating the distance between Na+ and Cl- ions:
Distance = 2 * r

By substituting the values of the atomic masses of Na and Cl, we can calculate the distance between Na+ and Cl- ions.

The correct answer is option 'B' (248.5 pm).