What is the relationship here? This is an annoying and probably easy question. How does one solve and approach it? 

 A: If we call $a$ the area of the shaded part, we see that the area of the small circle is $4 a$, and that of the larger circle is $6 a$. So the ratio of the two areas is $\frac{6}{4} $ (that is, $\frac {3}{2})$, and since area of a circle is proportional to the square of its radius, answer (A) looks to me like the right one.
A: Let $r$ and $R$ be the radius of the small and big circle respectively. You have that
$$\frac 14 \pi r^2=\mathrm{shaded}\:\: \mathrm{area}=\frac 16 \pi R^2.$$
Thus
$$\frac{R^2}{r^2}=\frac 32,$$ or equivalently,
$$\frac{R}{r}=\sqrt{\frac 32}.$$
Now it should be easy to get which possibility is the correct one.
A: $$\frac{A_1}{A_2}=\frac{1/4}{1/6}=\frac{3}{2}$$
$$\frac{r_1^2\pi}{r_2^2\pi}=\frac{3}{2}$$
$$\sqrt{3}/\sqrt{2}$$
A: $$\begin{array}{c}
{A_R} = \pi {R^2},{\rm{  }}{A_r} = \pi {r^2}\\
\frac{1}{6}{A_R} = \frac{1}{4}{A_r}\\
{A_R} = \frac{3}{2}{A_r}\\
\pi {R^2} = \frac{3}{2}\pi {r^2}\\
{R^2} = \frac{3}{2}{r^2}\\
R = \frac{{\sqrt 3 }}{{\sqrt 2 }}r\\
R:r = \sqrt 3 :\sqrt 2 
\end{array}$$
