Given, x + 2 ≤ $\frac{x}{3} + 3$ and x is a prime number. The solution set for x is :

Solving the given equation :

$$\begin{aligned}\Rightarrow x + 2 \le \frac{x}{3} + 3 \\[1em] \Rightarrow x - \frac{x}{3} \le 3 - 2 \\[1em] \Rightarrow \frac{3x - x}{3} \le 1 \\[1em] \Rightarrow \frac{2x}{3} \le 1 \\[1em] \Rightarrow x \le \frac{3}{2} \\[1em] \Rightarrow x \le 1.5\end{aligned}$$

Since, x is a prime number less than 1.5

Solution set is empty.

Hence, Option 1 is the correct option.