Calculate the atomic radius of iridium atom
Iridium has a face-center cubic (FCC) structure.

You assume that each iridium sphere has a radius $r$. But the edge of a unit cell has the dimension $a$. The volume of the unit cell, which is a cube, is $a^3$. The relationship between $a%$ and $r$ is that $a = 2.83r$.



FOLLOWING THE TEACHER"S(?) WORK...(or how you copied it?)

The following step is ok...
$$V = (2.83r)^3\tag{1}$$

But the second step here is wrong. The 2.83 should be cubed too.
$$5.7 \times 10^{-23} = 2.83r^3\tag{2}$$

(3) is totally wrong if we accept (2). 
$$r = 1.34 \times 10^{-8}\tag{3}$$

From (2)
$$r = \sqrt[3]{\dfrac{5.7\times 10^{-23}}{2.83}} = 2.72 \times 10^{-8}\tag{4}$$



Starting over...

$$V = (2.83r)^3\tag{5}$$
$$5.7 \times 10^{-23} = (2.83r)^3\tag{6}$$
$$r = \dfrac{\sqrt[3]{5.7 \times 10^{-23}}}{2.83} = 1.36 \times 10^{-8}\tag{7}$$

After calculating the the volume of the unit cell, I'd solved for $a$ by taking the cube root, then divided by 2.83 to get $r$. 

You have to be careful with significant figures too. It is not $5.7\times10^{-23}$ but rather $5.70\times10^{-23}$. Using a calculator I'd carried 5 significant figures in all the calculations then rounded to 3 at the end. In my day I used a slide rule so everything was to 3 significant figures.