It's a lot of circles!

Let $a$ be a side of square $ABCD$ and $x$ be a side of square $AJP_{1}I$ .

Extending the diagram to $n$ congruent circles we obtain:

$\sqrt{2}a = \sqrt{2}x + (n - 1)x + \dfrac{x}{2} + \dfrac{x}{\sqrt{2}} \implies 4a = (6 + \sqrt{2}(2n - 1))x \implies$

$x = \dfrac{4a}{6 + \sqrt{2}(2n - 1)} = \dfrac{4a}{\sqrt{2}(3\sqrt{2} + (2n - 1))} =$ $\dfrac{2\sqrt{2}}{3\sqrt{2} + (2n - 1)}a$

$\implies A_{AJP_{1}I} = x^2 = \dfrac{8}{4n^2 + 4(3\sqrt{2} - 1)n + 19 - 6\sqrt{2}}a^2 \implies$

$\dfrac{A_{AJP_{1}I}}{A_{ABCD}} = \dfrac{8}{4n^2 + 4(3\sqrt{2} - 1)n + 19 - 6\sqrt{2}} = \dfrac{24 - 16\sqrt{2}}{9} = \dfrac{8(3 - 2\sqrt{2})}{9}$

After simplifying we obtain:

$4(3 - 2\sqrt{2})n^2 + 4(11\sqrt{2} - 15)n + 72 - 56\sqrt{2} = 0 \implies$

$n = \dfrac{-11\sqrt{2} + 15 \pm 3\sqrt{3 - 2\sqrt{2}}}{2(3 - 2\sqrt{2})}$

$(\sqrt{2} - 1)^2 = 3 - 2\sqrt{2} \implies n = \dfrac{-11\sqrt{2} + 15 \pm 3(\sqrt{2} - 1)}{2(3 - 2\sqrt{2)}}$

For $+$ we obtain $n = \dfrac{4(3 - 2\sqrt{2})}{2(3 - 2\sqrt{2})} = 2$

For $-$ we obtain: $n = \dfrac{9 - 7\sqrt{2}}{3 - 2\sqrt{2}} = (9 - 7\sqrt{2})(3 + 2\sqrt{2}) = -1 - 3\sqrt{2} < 0 \therefore$ dropping the negative irrational root we obtain $n = \boxed{2}$ .