GIVEN :-

Lattice system = FCC
Density = 2180 kg / m³
Atomic weight of Na, M₁ = 23 kg / k.mol
Atomic weight of Cl, M₂ = 35.5 kg / k.mol
N = 6.02 x 10²⁶ atoms / k. mol.
For FCC, n = 4 atoms

Density = nM / N (a)³
And weight of NaCl molecule = M₁ + M₂
So, a³ = n (M₁ + M₂) / (Density) N
i.e. a³ = [(4 atoms) (23 + 35.5) kg / k.mol] / [(2180 kg / m³) (6.02 x 10²⁶ atoms / k. mol)]
i.e. a³ = 234 x 10^-31 m³ / 1.31
i.e. a³ = 178.63 x 10^-30 m³
So, Lattice constant, a = cube root of [178.63 x 10^-30 m³]
i.e. a = 5.63 A⁰

So the distance between adjacent atoms, sodium and chlorine is, d = a /2
i.e. d = (5.63 / 2) A⁰ = 2.815 A⁰