To find the set \( A \) defined as \( A = \{ x \in \mathbb{N} \mid \sqrt{x} \leq 4 \text{ and } x \text{ là bội của 3} \} \), we need to identify all natural numbers \( x \) that satisfy both conditions:

1. \( \sqrt{x} \leq 4 \)
2. \( x \) is a multiple of 3

Let's break down the conditions:

1. \( \sqrt{x} \leq 4 \)

This implies \( x \leq 4^2 = 16 \). So, \( x \) must be a natural number less than or equal to 16.

2. \( x \) is a multiple of 3

This means \( x \) must be in the form \( 3k \) where \( k \) is a natural number.

Now, let's list the natural numbers \( x \) that are multiples of 3 and less than or equal to 16:

- \( 3 \times 1 = 3 \)
- \( 3 \times 2 = 6 \)
- \( 3 \times 3 = 9 \)
- \( 3 \times 4 = 12 \)
- \( 3 \times 5 = 15 \)

So, the set \( A \) is:

\[ A = \{ 3, 6, 9, 12, 15 \} \]