An infinite rectangular stripe of width $3$ cm is folded along a line. What is the minimum possible area of the region of overlapping?
1. **Define the problem and setup the geometry:**
   - We have an infinite rectangular stripe of width \(3\) cm.
   - The stripe is folded along a line, creating a region of overlap.
   - We need to find the minimum possible area of this overlapping region.

2. **Identify key points and angles:**
   - Let \(A\) be a point on one edge of the stripe.
   - Let \(B\) and \(C\) be points on the other edge of the stripe such that \(\overline{BC}\) is the crease of the fold.
   - Define \(X\) as the midpoint of \(\overline{BC}\).
   - Let \(\angle CXY = \theta\) where \(0^\circ < \theta < 90^\circ\).

3. **Use geometric properties and trigonometric identities:**
   - Since \(\triangle ABX \cong \triangle ACX\) and \(\triangle ACX \sim \triangle CXY\), we can use these properties to find lengths in terms of \(\theta\).
   - The length \(CY\) can be expressed as:
     \[
     CY = XY \tan \theta = 1.5 \tan \theta
     \]
   - The length \(AY\) can be expressed as:
     \[
     AY = XY \cot \theta = 1.5 \cot \theta
     \]

4. **Calculate the area of the overlapping region:**
   - The area of \(\triangle ABC\) is twice the area of \(\triangle AXC\):
     \[
     [ABC] = 2 \cdot [AXC] = 2 \cdot \frac{XY \cdot AC}{2} = XY \cdot AC
     \]
   - Since \(AC = AY + CY\), we have:
     \[
     AC = 1.5 \tan \theta + 1.5 \cot \theta
     \]
   - Therefore, the area of \(\triangle ABC\) is:
     \[
     [ABC] = XY \cdot (AY + CY) = 1.5 \left(1.5 \tan \theta + 1.5 \cot \theta\right) = 2.25 \left(\tan \theta + \frac{1}{\tan \theta}\right)
     \]

5. **Minimize the expression using AM-GM inequality:**
   - By the AM-GM inequality, we know that:
     \[
     \tan \theta + \frac{1}{\tan \theta} \geq 2
     \]
   - Therefore:
     \[
     2.25 \left(\tan \theta + \frac{1}{\tan \theta}\right) \geq 2.25 \cdot 2 = 4.5
     \]
   - Equality is attained when \(\tan \theta = 1\), which implies \(\theta = 45^\circ\).

Conclusion:
\[
\boxed{4.5 \text{ cm}^2}
\]