\[ \text{density} = \frac{f}{\text{class width}} \]

\[ \frac{5}{4-0}=1.25,\quad \frac{9}{14-4}=0.9,\quad \frac{8}{20-14}=1.33. \]

The highest frequency density is \(1.33\), which corresponds to the interval \(14–20\).

\[ \text{density} = \frac{f}{\text{class width}} \]

\[ \frac{66}{6.5-2.5}=16.5,\quad \frac{80}{11.5-6.5}=16,\quad \frac{134}{20.5-11.5}=14.9. \]

The highest frequency density is \(16.5\), which corresponds to the interval \(3–6\).

Step 1: Convert density ratio to frequency ratio for the first and third classes.

Frequency \(=\) (frequency density) \(\times\) (class width). Widths: \(0\!-\!4\) has width \(4\); \(10\!-\!18\) has width \(8\).

\[ \text{freq ratio} \;=\; (4\times 4):(3\times 8) \;=\; 16:24 \;=\; 2:3. \]

Step 2: Split total frequency \(120\) in the ratio \(2:3\).

\[ \text{First class freq}=\frac{2}{2+3}\times 120=48, \qquad \text{Third class freq}=\frac{3}{2+3}\times 120=72. \]

Step 3: Find their densities.

\[ \text{First density}=\frac{48}{4}=12, \qquad \text{Third density}=\frac{72}{8}=9. \]

Step 4: Ensure the middle class is modal.

For \(4\!-\!10\) (width \(6\)) to be the modal class, its density must exceed the greatest of the other two densities, i.e. \(>12\).

\[ \text{Required freq} \;>\; 6\times 12 \;=\; 72. \]

The least possible integer frequency is therefore \(73\).