GIVEN :-

r = 1.746 A⁰ = 1.746 x 10^-10 m.
For FCC, a = 2 (square root of 2) r
i.e. a = 2 x 1.414 x 1.746 x 10^-10 m
i.e. a = 4.94 x 10^-10m

Spacing of (hkl) plane = d_hkl = [a / square root of ( h² + k² + l² )]

So, spacing of (202) plane = d₂₀₂ = [ 4.94 x 10^-10 / square root of ( 2² + 0² + 2² )]
i.e. d₂₀₂ = 1.75 x 10^-10m.

So, spacing of (220) plane = d₂₂₀ = [ 4.94 x 10^-10 / square root of ( 2² + 2² + 0² )]
i.e. d₂₂₀ = 1.75 x 10^-10m.

So, spacing of (111) plane = d₁₁₁ = [ 4.94 x 10^-10 / square root of ( 1² + 1² + 1² )]
i.e. d₁₁₁ = 1.01 x 10^-10m.