Let $ ABC$ be a triangle with circumradius $ R$, perimeter $ P$ and area $ K$. Determine the maximum value of: $ \frac{KP}{R^3}$.
1. **Establishing the inequality \(3\sqrt{3}R \ge P\):**
   - We start by noting that the perimeter \(P\) of the triangle can be expressed in terms of the sides \(a, b, c\) as \(P = a + b + c\).
   - Using the sine rule, we have \(a = 2R \sin A\), \(b = 2R \sin B\), and \(c = 2R \sin C\).
   - Therefore, \(P = 2R (\sin A + \sin B + \sin C)\).
   - To show \(3\sqrt{3}R \ge P\), it suffices to show that \(\sin A + \sin B + \sin C \le \frac{3\sqrt{3}}{2}\).
   - By Jensen's Inequality for the concave function \(\sin x\) on \([0, \pi]\), we have:
     \[
     \sin A + \sin B + \sin C \le 3 \sin \left(\frac{A + B + C}{3}\right) = 3 \sin \left(\frac{\pi}{3}\right) = 3 \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}.
     \]
   - Thus, \(P \le 3\sqrt{3}R\).

2. **Relating \(KP\) and \(R\):**
   - The area \(K\) of the triangle can be expressed using Heron's formula or the formula \(K = \frac{abc}{4R}\).
   - We need to show that:
     \[
     4\sqrt{3}KP = \frac{\sqrt{3}abcP}{R} \le 9abc.
     \]
   - Given \(P \le 3\sqrt{3}R\), we can write:
     \[
     \frac{\sqrt{3}abcP}{R} \le \frac{\sqrt{3}abc \cdot 3\sqrt{3}R}{R} = 9abc.
     \]
   - Therefore, \(4\sqrt{3}KP \le 9abc\), which simplifies to:
     \[
     KP \le \frac{9abc}{4\sqrt{3}}.
     \]

3. **Maximizing \(\frac{KP}{R^3}\):**
   - We need to show that:
     \[
     \frac{KP}{R^3} \le \frac{27}{4}.
     \]
   - From the previous step, we have:
     \[
     KP \le \frac{9abc}{4\sqrt{3}}.
     \]
   - Also, using the inequality \(3\sqrt{3}R \ge P\), we get:
     \[
     \frac{1}{R^3} \le \frac{3\sqrt{3}}{abc}.
     \]
   - Combining these, we have:
     \[
     \frac{KP}{R^3} \le \frac{9abc}{4\sqrt{3}} \cdot \frac{3\sqrt{3}}{abc} = \frac{27}{4}.
     \]

The final answer is \(\boxed{\frac{27}{4}}\).