**Solution:**

To find the volume of the solid of revolution, we need to integrate the cross-sectional area of the solid along the x-axis.

**Step 1: Finding the limits of integration**
To determine the limits of integration, we need to find the x-values where the curve intersects the x-axis. Setting y=0 in the given equation, we get:

0 = x^4 (x - 2)

Solving this equation, we find two solutions: x = 0 and x = 2. Therefore, the limits of integration are from x = 0 to x = 2.

**Step 2: Expressing the curve in terms of y**
Given equation: y^2 = x^4 (x - 2)

Simplifying the equation, we get:

y^2 = x^5 - 2x^4

Taking the square root of both sides, we have:

y = sqrt(x^5 - 2x^4)

**Step 3: Expressing the cross-sectional area**
The cross-sectional area of the solid at any x-value is given by the equation:

A(x) = π * R(x)^2

where R(x) is the radius of the solid at x. In this case, the radius is the y-value of the curve.

Therefore, the cross-sectional area is:

A(x) = π * (y)^2

Substituting the expression for y from Step 2, we have:

A(x) = π * (sqrt(x^5 - 2x^4))^2

Simplifying further, we get:

A(x) = π * (x^5 - 2x^4)

**Step 4: Integrating the cross-sectional area**
The volume of the solid is given by the integral of the cross-sectional area over the limits of integration:

V = ∫[0 to 2] A(x) dx

V = ∫[0 to 2] (π * (x^5 - 2x^4)) dx

Integrating, we get:

V = π * ∫[0 to 2] (x^5 - 2x^4) dx

V = π * [(1/6)x^6 - (2/5)x^5] [0 to 2]

V = π * [(1/6)(2)^6 - (2/5)(2)^5]

V = π * [(1/6)(64) - (2/5)(32)]

V = π * (10.67 - 12.8)

V ≈ -6.13π

Since volume cannot be negative, we take the absolute value:

|V| ≈ 6.13π

**Step 5: Rounding off the answer**
To round off the volume to 2 decimal places, we multiply the absolute value of the volume by π and round it to 2 decimal places.

Volume ≈ 6.13π ≈ 19.25 (rounded to 2 decimal places)

Since the given answer range is '6.60,6.80', it does not match the calculated volume. Therefore, the given answer is incorrect.