I was delaying with this question as seen below.
Calculate the radius of iridium atom, given that Ir has an FFC crystal structure, a density of $\pu{22.4 g/cm^3}$ and atomic weight of $\pu{192.2 g/mol}$.
Let me show you how my teacher solved this question.
Givendata:
$$\text {Weight of crystal} = \pu{192.2 g/cm^3}$$ $$\text {Density} = \pu{22.4 g/cm^3}$$ $$\text {Volume of crystal} = V = \frac {192.2}{22.4}$$ $$V = \pu{8.58 cm^3/mol}$$
In FFC, there are 4 atoms in one unit cell.
$$\text {Volume of one cell} = 8.58 \times \frac {4}{6.023 \times 10^{23}}$$ $$ = 5.7 \times 10^{-23}$$ $$V = a^3$$
$$\therefore a = 2.83r$$ and $$V = (2.83r)^3$$ $$5.7 \times 10^{-23} = 2.83r^3$$ $$r = 1.34 \times 10^{-8}$$
Where did that $a = 2.83r$ come from? That's what I didn't get.