To determine which statement(s) regarding the wire bent into a rectangle is/are correct, we need to go through a systematic analysis:

1. Understanding the Problem: We have a wire of length 20 cm, which is to be bent into a rectangle. We need to assess the truth of two statements regarding the resultant rectangle:
   • I. The rectangle of the largest area is the square.
   • II. It is possible to form a rectangle with an area of \(27 \, \text{cm}^2\).
2. Analyzing Statement I:
   • To maximize the area of a rectangle with a fixed perimeter, the rectangle should be a square. This is a derived fact from optimization where for a given perimeter, a square encloses the maximum area.
   • The perimeter of a square with side length \(s\) is \(4s\). Setting this equal to 20 cm gives:
   • The area of the square is \(s^2 = 5^2 = 25 \, \text{cm}^2\).
   • This confirms that Statement I is correct, as no rectangle of the same perimeter can have an area larger than 25 cm².
3. Analyzing Statement II:
   • Let's denote the length of the rectangle as \(l\) and width as \(w\).
   • We know the perimeter is given by \(2l + 2w = 20\) which simplifies to \(l + w = 10\).
   • The area \(A\) of the rectangle is \(lw\). We need this area to be 27 cm².
   • Substitute \(w = 10 - l\) into the area equation:
   • The discriminant of this quadratic equation is \(b^2 - 4ac = (-10)^2 - 4 \times 1 \times 27 = 100 - 108 = -8\).
   • The negative discriminant indicates that no real solution exists for \(l\) and \(w\) that would satisfy the equation. Thus, Statement II is incorrect.
4. Conclusion: Only Statement I is correct, as the largest area for the given perimeter configuration is indeed a square of 25 cm². Statement II is not feasible due to the negative discriminant in the quadratic equation, showing no real rectangle can have an area of 27 cm² with the given perimeter.