The elements given are

Li, Na, Be, Mg, B and Al

2$^{nd}$ period elements $\to$ Li, Be, B

Order of atomic radii $\to$ Li > Be > B

3$^{rd}$ period elements $\to$ Na, Mg, Al

Order of atomic radii $\to$ Na > Mg > Al

In a period, atomic radii decreases on moving from left to right.

So, the element with least atomic radii in 2$_{nd}$ period (Li, Be, B) is B. The element with least atomic radii 3$_{rd}$ period (Na, Mg, Al) is Al.

In B and Al $\Rightarrow$ B and Al are same group elements. Down the group, radii increases. So, B has least radii.

B is the element with least atomic radius to form the oxide $A_2O_3$.

Oxidation states of the A in oxides :

$${A_0} \to + 2\,\,{A_{{0_2}}} \to + 2\,\,{A_{{2^0}}} \to + 1\,\,{A_{2_3^0}} \to + 3$$

B outer configuration is s$^2$P$^1$. 3 valence electrons and form A$_2$O$_3$ type oxide.