Step 1: Recall the density formula.
For a cubic crystal, density is \[ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \] where $Z$ is atoms per unit cell, $M$ is molar mass, $a$ is edge length and $N_A$ is Avogadro number.
Step 2: List the data.
Here $M = 75$, $a = 5$ angstrom which is $5 \times 10^{-8}$ cm, $\rho = 2$ and $N_A = 6 \times 10^{23}$.
Step 3: Solve for $Z$.
Rearrange to find $Z$. \[ Z = \frac{\rho \, a^3 \, N_A}{M} = \frac{2 \times (5\times10^{-8})^3 \times 6\times10^{23}}{75} \]
Step 4: Work out the number.
Since $(5\times10^{-8})^3 = 125 \times 10^{-24}$, the top becomes $2 \times 125 \times 10^{-24} \times 6 \times 10^{23} = 150$. Divide by 75 to get $Z = 2$.
Step 5: Identify the lattice and radius rule.
Two atoms per cell means a body-centred cubic lattice. For BCC the radius and edge are linked by $4r = \sqrt{3}\,a$, so $r = \dfrac{\sqrt{3}}{4}a$.
Step 6: Find the radius.
Put $a = 5$ and $\sqrt{3} = 1.732$. \[ r = \frac{1.732 \times 5}{4} = \frac{8.66}{4} = 2.165\ \text{angstrom} \] \[ \boxed{r = 2.165\ \text{angstrom}} \]