(A) Yes,it is true.
Let the side of the square be $a = 5 \, cm$.
The diameter of the inner circle is equal to the side of the square,so $d_{inner} = a = 5 \, cm$. The radius is $r = \frac{5}{2} \, cm$.
The area of the inner circle is $A_{inner} = \pi r^2 = \pi \left(\frac{5}{2}\right)^2 = \frac{25\pi}{4} \, cm^2$.
The diameter of the outer circle is equal to the diagonal of the square,so $d_{outer} = a\sqrt{2} = 5\sqrt{2} \, cm$. The radius is $R = \frac{5\sqrt{2}}{2} \, cm$.
The area of the outer circle is $A_{outer} = \pi R^2 = \pi \left(\frac{5\sqrt{2}}{2}\right)^2 = \pi \left(\frac{25 \times 2}{4}\right) = \frac{50\pi}{4} = \frac{25\pi}{2} \, cm^2$.
Comparing the two areas: $\frac{A_{outer}}{A_{inner}} = \frac{25\pi / 2}{25\pi / 4} = \frac{4}{2} = 2$.
Thus,the area of the outer circle is indeed two times the area of the inner circle.