#### Answer

BCC

#### Work Step by Step

Note: $M$ is the atomic weight, $N_A$ is the Avogadro's number, $\rho$ is the density
(BCC): $8 \times 1/8$ atoms in the corners and one in the center, so 2 atoms per unit cell;
Cube's diagonal $l.\sqrt3=4r\rightarrow l=4r/\sqrt3; V=l^3$
$\rho=(2\times M/N_A)\div(64.r^3/3\sqrt3)=(3\sqrt3/32)\times(\dfrac{M}{N_A.r^3})$
(FCC) $8 \times 1/8$ atoms in the corners and $6\times1/2$ in the faces, 4 atoms per unit cell
Face's diagonal: $l.\sqrt2=4.r\rightarrow l=4r/\sqrt2; V=l^3$
$\rho=(4\times M/N_A)\div(64.r^3/2\sqrt2)=(\sqrt2/8)\times(\dfrac{M}{N_A.r^3})$
(Primitive) $8\times 1/8$ one atom per cell, $l=2.r$
$\rho=(M/N_A)\div(8.r^3)=(1/8)\times(\dfrac{M}{N_A.r^3})$
The factor $\dfrac{M}{N_A.r^3}$ is $50.9415/6.022\times10^{23}.(132\times10^{-10})^3=36.78\ g/cm^3$
The density divided by this factor is $6.11/36.78=0.166$ which is closer to $3\sqrt3/32$, so the unit cell is in a body-centered cubic arrangement.