$\begin{array}{1 1}152.4pm\\162.4pm\\142.4pm\\132.4pm\end{array} $

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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$

$\;\;=152.4pm$

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