To find the volume of the solid of revolution, we use the formula:

\(V = \pi \int_{a}^{b} [f(x)]^2 \, dx\)

Given \(f(x) = \frac{3}{2x+5}\), we have:

\([f(x)]^2 = \left(\frac{3}{2x+5}\right)^2 = \frac{9}{(2x+5)^2}\)

The limits of integration are from \(x = 0\) to \(x = 2\).

Thus, the volume \(V\) is:

\(V = \pi \int_{0}^{2} \frac{9}{(2x+5)^2} \, dx\)

Integrating, we have:

\(V = \pi \left[ -\frac{9}{2(2x+5)} \right]_{0}^{2}\)

Evaluating the definite integral:

\(V = \pi \left( -\frac{9}{2(2 \times 2 + 5)} + \frac{9}{2(2 \times 0 + 5)} \right)\)

\(V = \pi \left( -\frac{9}{18} + \frac{9}{10} \right)\)

\(V = \pi \left( -0.5 + 0.9 \right)\)

\(V = \pi \times 0.4\)

\(V = 0.4\pi\)

Thus, the volume is \(0.4\pi\) (or approximately 1.26).