To solve this problem, let's break it down step by step.

Step 1: Side lengths of the squares

• The perimeter of the smaller square is given as $40$. Since the perimeter of a square is four times the length of one side, the side length of the smaller square $s_1$ is:

$s_1 = \frac{40}{4} = 10$

• The perimeter of the larger square is given as $56$. Similarly, the side length of the larger square $s_2$ is:

$s_2 = \frac{56}{4} = 14$

Step 2: Geometry of the inscribed square

The smaller square is inscribed in the larger square, meaning the vertices of the smaller square touch the sides of the larger square. In such an arrangement, the smaller square is rotated by 45° relative to the larger square.

The diagonal of the smaller square is aligned with the sides of the larger square. The diagonal of a square can be calculated using the formula:

$\text{Diagonal} = s \sqrt{2}$

For the smaller square:

$\text{Diagonal of the smaller square} = s_1 \sqrt{2} = 10 \sqrt{2}$

This diagonal is equal to the side length of the larger square. Therefore:

$10 \sqrt{2} = 14$

Step 3: Calculate the greatest distance

We are asked to find the greatest distance between a vertex of the inner square and a vertex of the outer square, denoted as $r \sqrt{abc}$, where $abc$ is a three-digit natural number.

From the geometry of the inscribed squares, the greatest distance occurs between the vertices where the inner square meets the midpoint of the sides of the outer square. By using the properties of the diagonals and positioning,