Answer : $7.30gcm^{-3}$
In a body-centred cubic lattice ,atoms touch each other along the body-diagonal of the cube.Hence
$4r=\sqrt 3a$ or $r=\large\frac{\sqrt 3 a}{4}=\frac{(1.732)(287 pm)}{4}=$$124.27$pm.
The expression of density is $\rho=\large\frac{N}{a^3}\big(\frac{M}{N_A}\big)$
Substituting the values,we get
$\rho=\large\frac{2}{(287\times 10^{-10}cm)^3}\big(\frac{51.99gmol^{-1}}{6.023\times 10^{23}mol^{-1}}\big)$$=7.30gcm^{-3}$