Problem 1245

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 $\boxed{\frac{27}{4}}$.