Comparing Areas

Let the side length of the large square be $1$ , then the diameter of the large circle is $1$ .

Applying Pythagorean theorem on the big circle, the side length of the small square is $\sqrt{\dfrac{1}{2}}$ , then the diameter of the small circle is also $\sqrt{\dfrac{1}{2}}$ .

$A.$ $\dfrac{1^2}{\left(\sqrt{\dfrac{1}{2}}\right)^2}=\dfrac{1}{\frac{1}{2}}=1 \times 2=$ $\color{plum}\large \boxed{2}$

$B.$ $\dfrac{2\left(\sqrt{\dfrac{1}{2}}\right)^2}{1^2}=\dfrac{2 \times \frac{1}{2}}{1}=\dfrac{1}{1}=$ $\color{plum} \large \boxed{1}$

$\color{#D61F06}\boxed{\large Conlusion: A>B}$