Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 $$\mathring{A}$$. The radius of sodium atom is approximately:

We are told that sodium crystallizes in a body-centred cubic (bcc) lattice. In a bcc lattice each corner atom touches the single atom at the body centre along the body diagonal of the cube.

The length of the body diagonal of a cube of edge $$a$$ is, by the Pythagoras theorem, $$\sqrt{3}\,a$$.

Along that diagonal we have one corner atom, then the body-centred atom, and finally the opposite corner atom. The sequence of radii is therefore $$r + 2r + r = 4r$$. Hence for a bcc lattice

$$\sqrt{3}\,a = 4r.$$

The edge length is given as $$a = 4.29\;\text{\AA}$$. Substituting this value, we obtain

$$4r = \sqrt{3}\times 4.29\;\text{\AA}.$$

Using $$\sqrt{3}\approx 1.732$$, we have

$$4r = 1.732 \times 4.29\;\text{\AA}.$$

Carrying out the multiplication step by step:

$$1.732 \times 4 = 6.928,$$

$$1.732 \times 0.29 \approx 0.502,$$

so

$$1.732 \times 4.29 \approx 6.928 + 0.502 = 7.430.$$

Thus

$$4r \approx 7.430\;\text{\AA}.$$

Dividing both sides by 4,

$$r = \frac{7.430}{4}\;\text{\AA}.$$

$$r \approx 1.8575\;\text{\AA}.$$

Rounded to three significant figures,

$$r \approx 1.86\;\text{\AA}.$$

This value corresponds to the second option in the list.

Hence, the correct answer is Option B.