In a bcc unit cell,the atoms touch each other along the body diagonal,so,if the radius of the atom is r,then
Body diagonal =r+2r+r=4r
From the geometry of a cube
Body diagonal =$\sqrt 3\times $edge length=$\sqrt 3a$
Then,$4r=\sqrt 3a$
$\therefore$ Radius of metal atom =$\large\frac{\sqrt 3}{4}$$a$
$r=\large\frac{\sqrt 3}{4}$$\times a=\large\frac{\sqrt 3}{4}$$\times 352$
$\;\;=152.4pm$