Lithium forms body centred cubic structure. The length of the side of its unit cell is $$351$$ pm. Atomic radius of the lithium will be:

For lithium the crystal structure is body-centred cubic (bcc).

In a bcc unit cell the atoms touch each other along the body diagonal.
Length of the body diagonal = $$\sqrt{3}\,a$$, where $$a$$ is the edge length.
Along this diagonal we have: corner atom (radius $$r$$) + body-centre atom (diameter $$2r$$) + opposite corner atom (radius $$r$$).
Therefore

$$\sqrt{3}\,a = 4r \qquad -(1)$$

Given $$a = 351\;\text{pm}$$, substitute in equation (1):

$$r = \frac{\sqrt{3}}{4}\,a = \frac{1.732}{4} \times 351\;\text{pm}$$

$$r = 0.433 \times 351\;\text{pm} \approx 151.9\;\text{pm}$$

Rounded to the nearest whole number, the atomic radius of lithium is $$152\;\text{pm}$$.

Hence, Option D which is: $$152\;\text{pm}$$.