The type of oxide formed by the element among Li, Na, Be, Mg, B and Al that has the least atomic radius is:

🧠 Smallest atomic radius among Li, Na, Be, Mg, B, Al → Boron. Boron's oxide is $\ce{B2O3}$, of type $\text{A}_2\text{O}_3$ You first need to identify which element has the smallest radius, then write its oxide formula.

🗺️ Step 1 — find the smallest atom Period 2 atoms (Li, Be, B) are smaller than Period 3 atoms (Na, Mg, Al). So eliminate Na, Mg, Al. Within Period 2: radius decreases left-to-right. Li ($152$ pm) > Be ($112$ pm) > B ($88$ pm). So Boron has the smallest radius.

Step 2 — Boron's oxide B is in Group 13 with valency 3. With oxygen (valency 2), criss-cross to get $\ce{B2O3}$. Type: $\text{A}_2\text{O}_3$.

⚠️ Common mistake Some students pick option (d) $\text{A}_2\text{O}$ thinking the smallest atom must be Lithium (since Li is at the very top of Group 1). But B sits to the right of Li in Period 2, with three more protons — so B's atom is smaller, not Li's. Always check both period AND group when comparing atomic radii across non-adjacent elements.

$\boxed{\text{Answer: (a)}}$