For convenience, we assume the $a_{i}$ are indexed modulo 101, so that $a_{i+1}=a_{1}$ when $a_{i}=a_{101}$. In any consistent subset of $C_{101}$ of order 1, $b$ must be paired with exactly one $a_{i}$, say $a_{1}$. Then, $a_{2}$ cannot be paired with $a_{1}$, so it must be paired with $a_{3}$, and likewise we find we use the pairs $\{a_{4}, a_{5}\}, \{a_{6}, a_{7}\}, \ldots, \{a_{100}, a_{101}\}$ - and this does give us a consistent subset of order 1. Similarly, pairing $b$ with any other $a_{i}$ would give us a unique extension to a consistent configuration of order 1. Thus, we have one such 2-configuration for each $i$, giving $S_{1}(101)=101$ altogether. In a consistent subset of order 2, $b$ must be paired with two other elements. Suppose one of them is $a_{i}$. Then $a_{i}$ is also paired with either $a_{i-1}$ or $a_{i+1}$, say $a_{i+1}$. But then $a_{i-1}$ needs to be paired up with two other elements, and $a_{i}$ is not available, so it must be paired with $a_{i-2}$ and $b$. Now $b$ has its two pairs determined, so nothing else can be paired with $b$. Thus, for $j \neq i-1, i$, we have that $a_{j}$ must be paired with $a_{j-1}$ and $a_{j+1}$. So our subset must be of the form $\{\{b, a_{i}\}, \{a_{i}, a_{i+1}\}, \{a_{i+1}, a_{i+2}\}, \ldots, \{a_{101}, a_{1}\}, \ldots, \{a_{i-2}, a_{i-1}\}, \{a_{i-1}, b\}\}$ for some $i$. On the other hand, for any $i=1, \ldots, 101$, this gives a subset meeting our requirements. So, we have 101 possibilities, and $S_{2}(101)=101$. Finally, in a consistent subset of order 3, each $a_{i}$ must be paired with $a_{i-1}, a_{i+1}$, and $b$. But then $b$ occurs in 101 pairs, not just 3, so we have a contradiction. Thus, no such subset exists, so $S_{3}(101)=0$.