Official Solution: For the upper bound, to each arrangement assign a string of parentheses. First arrange the circles in such a way that each circle's center is on the x axis, and then cut the tops and bottoms of every circle. The remaining of each circle consists of two arcs shaped either like $ ($ or like $ )$. Assign this string to this arrangement. (Of course there may be several ways to assign a string to a single arrangement) This string is correctly parenthesized, because each $ ($ is matched to a corresponding $ )$, and the substring between these two, corresponds to a subarrangement with fewer circles (that is those circles contained in one of the circles in the original arrangement) It is straightforward to construct back the arrangement from the string of parentheses (Complete the circles by joining each $ ($ with its matching $ )$) Therefor $ a_n$ is smaller than or equal to the Catalan number of order $ n$ which is $ \frac 1{n+1}{2n\choose n}\leq 2^{2n}=4^n$. For the lower bound, select a string of $ 0$'s and $ 1$'s of length $ n$ and assign an arrangement to it. Reading the string from left to right, replace each $ 0$ by a circle containing no other circles, and each $ 1$ by a circle containing all the other previous circles. Of course some arrangements can not be achieved by this method. But if an arrangement can be achieved, it can be achieved from at most one string. This is because, we can easily reverse the procedure. first consider the outermost circles; at most one can contain other circles. Put as many $ 0$'s as there are empty circles at the end of the string. If no circle is left, then we are done. If there is one left, put a $ 1$ before these zeros, and then put the string representing the contents of this circle at the beginning of our string. This shows that $ a_n\geq 2^n$.

Here's a different way to get the upper bound for (a). Set $a_0 = 1$ for convenience. Call a circle in a good set tasty if it is not contained by any other circles. Define the size of a tasty circle as the number of circles which are inside it (including itself). By doing casework on the sizes of the tasty circles, we derive the following recurrence relation for all $n \ge 1$: $$a_n = \sum_{x_1, x_2, \cdots, x_k \in \mathbb{N}, x_1 + x_2 + \cdots + x_k = n} a_{x_1 - 1} a_{x_2 - 1} \cdots a_{x_k - 1},$$ where the sum is over all unordered partitions $(x_1, x_2, \cdots, x_k)$ of $n$. It's clear that $a_n$ only increases for all $n$ if we instead sum over ordered partitions $(x_1, x_2, \cdots, x_k)$ of $n$, so let $b_n$ be the set which follows this recurrence relation instead. In other words, we have $b_0 = 1, b_1 = 1, b_2 = 2, b_3 = 5, \cdots$ and $$b_n = \sum_{x_1, x_2, \cdots, x_k \in \mathbb{N}, x_1 + x_2 + \cdots + x_k = n} b_{x_1 - 1} b_{x_2 - 1} \cdots b_{x_k - 1},$$ where the sum is over all ordered partitions $(x_1,x_2, \cdots, x_k)$ of $n$. Let $M(x) = \sum_{k = 1}^{\infty} b_k x^k.$ Then the above recurrence relation implies that: $$M(x) = \frac{x + xM(x)}{1 - x - xM(x)}.$$ Expanding, we get $xM(x)^2 + (2x-1) M(x) + x = 0$, and solving this quadratic gives: $$M(x) = \frac{-(2x-1) \pm \sqrt{1 - 4x}}{2x}.$$ By plugging in $x = 0$, it's clear that because $M(0) = 1$, we must have a $-$ instead of a $+$, so that: $$M(x) = \frac{-(2x-1) - \sqrt{1-4x}}{2x}.$$ By the Binomial Theorem, we then have that the $x^n-$coefficient in $M(x)$ is $\frac{\binom{2n}{n}}{2(2n-1)}$ for $n \ge 1.$ This means that $b_n \le \frac{\binom{2n}{n}}{2(2n-1)}$ for $n \ge 1$, and so clearly the growth ratio of $\{b_n\}$ is smaller than $4.$ Since $a_n \le b_n$ for all $n \in \mathbb{N}$, so is the growth ratio of $\{a_n\}.$ $\square$ P.S. Would like to see a way to improve the upper bound on the growth rate, because this solution bounds $\{a_n\}$ very loosely.