Three. (20 points) As shown in Figure 1, square $A B C D$ is inscribed in an ellipse, and squares $E F G H$ and $I J H K$ have vertices $E$, $F$, and $I$ on the ellipse, vertices $K$, $H$, and $G$ on side $A B$, and vertex $J$ on side $H E$. It is known that the area ratio of square $A B C D$ to square $E F G H$ is $4: 1$. Find the area ratio of square $I J H K$ to square $E F G H$ (accurate to 0.001).
Three, let the equation of the ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, and the side length of the square $A B C D$ be $2 \mathrm{~m}$. Then
$A(m, m), E\left(2 m, \frac{m}{2}\right)$.
Thus, $\left\{\begin{array}{l}\frac{m^{2}}{a^{2}}+\frac{m^{2}}{b^{2}}=1, \\ \frac{4 m^{2}}{a^{2}}+\frac{m^{2}}{4 b^{2}}=1 .\end{array}\right.$
(1) $\times 4$ - (2) to eliminate $a$ gives $b^{2}=\frac{5 m^{2}}{4}$.
(2) $\times 4$ - (1) to eliminate $b$ gives $a^{2}=5 m^{2}$.
Therefore, the equation of the ellipse can be written as
$$
\frac{x^{2}}{5 m^{2}}+\frac{4 y^{2}}{5 m^{2}}=1 \text {. }
$$

Let the side length of the square $I J H K$ be $t$. Then
$$
I\left(m+t, \frac{m}{2}+t\right) \text {. }
$$

Substituting the coordinates of point $I$ into equation (3) gives
$$
\frac{(m+t)^{2}}{5 m^{2}}+\frac{4\left(\frac{m}{2}+t\right)^{2}}{5 m^{2}}=1,
$$

which simplifies to $5 t^{2}+6 m t-3 m^{2}=0$.
Solving for $t$ gives $t=\frac{-3 \pm 2 \sqrt{6}}{5} m$ (the negative value is discarded).
Thus, the area of the square $I J H K$ is
$$
t^{2}=\left(\frac{-3+2 \sqrt{6}}{5} m\right)^{2}=0.1442 m^{2} \text {. }
$$

Therefore, the area ratio of square IJHK to square $E F G H$ is $0.144: 1$.