$$V=a^3$$

Where 'a' is the length of the edge of the cube.

The volume of one Niobium atom is:

$$V_{Nb}=(4/3)\pi r^3$$

Since there are two atoms per unit cell, the volume of two Niobium atoms is:

$$2V_{Nb}=(8/3)\pi r^3$$

Setting these two volumes equal to each other, we get:

$$a^3=(8/3)\pi r^3$$

Solving for 'r', we get:

$$r=\sqrt[3]{\frac{3a^3}{8\pi}}$$

The density of Niobium is given by:

$$\rho=\frac{2M}{a^3N_A}$$

Where M is the molar mass of Niobium (92.91 g/mol), $N_A$ is Avogadro's number (6.022 x 10^23 atoms/mol), and 'a' is the length of the edge of the cube.

Solving for 'a', we get:

$$a=\sqrt[3]{\frac{2M}{\rho N_A}}$$

Substituting this expression for 'a' into the equation for 'r', we get:

$$r=\sqrt[3]{\frac{3(2M/\rho N_A)^3}{8\pi}}$$

Plugging in the values for M, $\rho$, and $N_A$, we get:

$$r=\sqrt[3]{\frac{3(2\times92.91\text{ g/mol}/8.57\text{ g/cm}^3\times6.022\times10^{23}\text{ atoms/mol})^3}{8\pi}}$$

$$r=1.43\times10^{-8}\text{ cm}$$

Therefore, the radius of a Niobium atom is $$1.43\times10^{-8}\text{ cm}$$.