GIVEN :-

r = 1.44 A⁰ = 1.44 x 10^-10 m.
For FCC, a = 2 (square root of 2) r
i.e. a = 2 x 1.414 x 1.44 x 10^-10 m
i.e. a = 4.07 x 10^-10m

Spacing of (111) plane = d₁₁₁ = [a / square root of ( h² + k² + l² )]
i.e. d₁₁₁ = [ 4.07 x 10^-10 / square root of ( 1² + 1² + 1² )]
i.e. d₁₁₁ = 2.35 x 10^-10m.

We know that, 2d sin(theta) = n Lamda
i.e.2 x 2.35 x 10^-10 x sin (32.1) = 1 lamda
So, lamda = 2.498 x 10^-10m