Atomic Radius in Cubic Unit Cells

In a face centred cubic (fcc) crystal the atoms touch each other along the face diagonal.
Face diagonal $= \sqrt{2}\,a = 4r$, where $a$ is the lattice parameter.
$r = \dfrac{\sqrt{2}\,a}{4} = \dfrac{a}{2\sqrt{2}}$
$r = \dfrac{\sqrt{2} \times 3.6}{4}$
$r = 0.9 \times 1.41$
$r = 1.27$ Å

In an fcc lattice the atoms touch each other along the face diagonal: $4r = \sqrt{2}\,a$
The interatomic spacing (nearest neighbour distance) is $2r$.
So $2r = \dfrac{\sqrt{2}\,a}{2} = \dfrac{a}{\sqrt{2}}$
Given $2r = 2.54$ Å, so $a = \sqrt{2} \times 2.54$
$a = 1.414 \times 2.54$
$a = 3.59$ Å