A certain element crystallises in a body-centred cubic (bcc)...

For a body-centred cubic (bcc) lattice, the number of atoms per unit cell, $Z = 2$.

The mass of $5 \times 10^{23}$ atoms is $27 \text{ g}$.
Mass of one atom, $m = \dfrac{27}{5 \times 10^{23}} = 5.4 \times 10^{-23} \text{ g}$.

The density of the crystal is given by:
$\rho = \dfrac{Z \times m}{a^3}$

Substituting the given values:
$4.0 = \dfrac{2 \times 5.4 \times 10^{-23}}{a^3}$

$a^3 = \dfrac{10.8 \times 10^{-23}}{4.0} = 2.7 \times 10^{-23} \text{ cm}^3$

$a^3 = 27 \times 10^{-24} \text{ cm}^3$

Taking the cube root on both sides:
$a = 3 \times 10^{-8} \text{ cm}$

Converting to picometers ($1 \text{ cm} = 10^{10} \text{ pm}$):
$a = 3 \times 10^{-8} \times 10^{10} \text{ pm} = 300 \text{ pm}$.

Answer: $300$