is:

The correct option is 56√3ab.

Step 1: Identify the given equations from the image
From the provided image, the region is bounded by the following lines:
1. The line:

$$\frac{x}{7\sqrt{3}a}+\frac{y}{b}=4$$

2. The y-axis:

$$x=0$$

3. The x-axis:

$$y=0$$

Step 2: Convert the line's equation into the standard intercept form
The standard intercept form of a straight line is:

$$\frac{x}{A}+\frac{y}{B}=1$$

where $A$ is the x-intercept and $B$ is the y-intercept.
To convert the given equation to this form, divide both sides of the equation by 4:

$$\frac{x}{4\times 7\sqrt{3}a}+\frac{y}{4b}=1$$

This simplifies to:

$$\frac{x}{28\sqrt{3}a}+\frac{y}{4b}=1$$

Step 3: Determine the intercepts
Comparing the simplified equation with the intercept form, we obtain:
- Base (x-intercept, $A$):

$$A=28\sqrt{3}a$$

- Height (y-intercept, $B$):

$$B=4b$$

Step 4: Calculate the area of the region
The region bounded by the coordinate axes ($x=0$ and $y=0$) and the line forms a right-angled triangle with the vertex at the origin $(0,0)$.
The area of a right-angled triangle is given by:

$$\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$$

Substituting the values of base ($A$) and height ($B$):

$$\text{Area}=\frac{1}{2}\times \left(28\sqrt{3}a\right)\times \left(4b\right)$$

Simplifying the expression:

$$\text{Area}=14\sqrt{3}a\times 4b$$

$$\text{Area}=56\sqrt{3}ab$$

Thus, the area of the bounded region is 56√3ab.