The moment of inertia \( I \) for a rectangular cross section is given by the formula:
\[ I = \frac{1}{12} b h^3 \]
where \( b \) is the width and \( h \) is the height of the beam.
Let's denote:
- Beam 1: rectangular cross section, width = \( b \) and height = \( h \).
- Beam 2: composed of three thinner beams, width \( b \) and height \( \frac{h}{3} \).
Calculate \( I_1 \) for Beam 1:
For Beam 1, the moment of inertia \( I_1 \) is:
\[ I_1 = \frac{1}{12} b h^3 \]
Calculate \( I_2 \) for Beam 2:
Beam 2 consists of three beams, each having a height of \( \frac{h}{3} \). The width remains the same, and we'll find the moment of inertia of one of the thinner beams and then combine them.
For one thinner beam (height \( \frac{h}{3} \)), the moment of inertia \( I_{\text{thin}} \) is:
\[ I_{\text{thin}} = \frac{1}{12} b \left( \frac{h}{3} \right)^3 = \frac{1}{12} b \frac{h^3}{27} = \frac{b h^3}{324} \]
Since there are three such beams, the total moment of inertia \( I_2 \) is:
\[ I_2 = 3 I_{\text{thin}} = 3 \left( \frac{b h^3}{324} \right) = \frac{3b h^3}{324} = \frac{b h^3}{108} \]
Calculate the ratio \( \frac{I_1}{I_2} \):
Now, we can find the ratio \( \frac{I_1}{I_2} \):
\[ \frac{I_1}{I_2} = \frac{\frac{1}{12} b h^3}{\frac{b h^3}{108}} = \frac{1}{12} \cdot \frac{108}{1} = \frac{108}{12} = 9 \]
So the ratio of the moments of inertia \( I_1/I_2 \) is:
\[ \frac{I_1}{I_2} = 9 \]
Thus, the final answer is:
\[ \boxed{9} \]