Answer: D. $43.33$

Explanation:1. Original Map Details: Side length of the original square map = $55$ inches. Scale of the original map: $1$ inch on map represents $13$ miles in actual distance.

2. Smaller Map Details: The side length of the smaller map is $70\%$ shorter than the side length of the previous map. This means the new side length is $100\% - 70\% = 30\%$ of the original side length. Side length of smaller map = $30\%$ of $55$ inches Side length of smaller map = $0.30 \times 55$ inches Side length of smaller map = $16.5$ inches.

3. Actual Dimensions Represented by Both Maps: Both maps represent the same actual square area. Actual side length represented by the original map = $55 \text{ inches} \times 13 \text{ miles/inch}$ Actual side length represented = $715$ miles. So the area depicted is a square of $715$ miles by $715$ miles.

4. Scale of the Smaller Map: The smaller map has a side length of $16.5$ inches, and this $16.5$ inches represents an actual distance of $715$ miles. We want to find the actual distance, in miles, represented by $1$ inch on this smaller map. Let $s\_{new}$ be the new scale in miles per inch. $s\_{new} = \dfrac{\text{Actual distance}}{\text{Map distance (smaller map)}}$ $s\_{new} = \dfrac{715 \text{ miles}}{16.5 \text{ inches}}$

5. Calculate the New Scale: $s\_{new} = \dfrac{715}{16.5}$ To simplify the division, multiply numerator and denominator by $10$: $s\_{new} = \dfrac{7150}{165}$ Divide both by $5$: $s\_{new} = \dfrac{1430}{33}$ Divide both by $11$: $s\_{new} = \dfrac{130}{3}$ Now convert to decimal: $130 \div 3 = 43.333...$

6. Compare with Options: The value $43.333...$ miles per inch. [A] $3.90$ [B] $7.65$ [C] $38.50$ [D] $43.33$ - This is the closest to our calculated value.

7. Conclusion: On the smaller map, $1$ inch represents approximately $43.33$ miles.