Solution:
$\text{Across a period (left to right), atomic radii decrease due to increasing nuclear charge pulling electrons closer.}$ $\text{Down a group (top to bottom), radii increase as additional electron shells are added, outweighing the effect of nuclear charge.}$ $(1) \ Mg \ \text{(Period 3) and} \ Al \ \text{(Period 3) have larger radii than} \ C \ \text{(Period 2) and} \ O \ \text{(Period 2) due to having an additional electron shell.}$ $\text{In Period 3, radii decrease left to right as nuclear charge increases so atomic radii of} \ Mg>Al.$ $\text{In Period 2, radii similarly decrease from} \ C \ \text{to} \ O \ \text{with higher atomic number.}$ $\text{The correct order of atomic radii will be:} \ Mg > Al > C > O$ $(2) \ \text{Aluminum (Al) is in Group 13 and has a larger radius than elements in the same period due to its position in the periodic table.}$ $\text{Boron (B) is in Group 13 but in Period 2, making it smaller than Al but larger than elements in Period 2 like N and F.}$ $\text{Nitrogen (N) and Fluorine (F) are in Period 2, with N being larger than F due to F having a higher atomic number and thus a stronger nuclear pull on its electrons.}$ $\text{The correct order of atomic radii will be:} \ Al > B > N > F$ $(3) \ \text{Beryllium (Be) and Magnesium (Mg) belong to the same group (Group 2), but Mg is below Be, meaning Mg has a larger atomic radius.}$ $\text{Aluminum (Al) and Silicon (Si) are in the same period (Period 3), and moving from left to right, the atomic radius decreases, so Al has a larger atomic radius than Si.}$ $\text{Thus, the correct order of atomic radii will be:} \ Be < Mg > Al > Si$ $(4) \ \text{Fluorine has the smallest radius as it is furthest to the right in Period 2.}$ $\text{Chlorine, phosphorus, and silicon follow Period 3 trends, with silicon having the largest radius.}$ $\text{So, the correct order of atomic radii will be:} \ Si > P > Cl > F$ $\text{Hence, option 3 is the correct answer.}$