From the expression $\rho=\large\frac{n}{a^3}\big(\large\frac{M}{N_A}\big)$
We get $n=\large\frac{\rho a^3N_A}{M}=\frac{1.54\times (556\times 10^{-10})^3\times 6.023\times 10^{23}}{40.08}$
$\Rightarrow 3.98\simeq 4$
For n=4,the unit cell is face-centred cube.Since atoms touch each other along the face diagonal in a face-centred cubic unit cell,we have
$4r=\sqrt 2a$
$r=\large\frac{a}{\sqrt 2}$
$\;\;=\large\frac{556}{2\sqrt 2}$pm
$\;\;=196.6$pm