Solution:

The answer is (C). We distinguish several cases, depending on the number of prime factors of $b$:

a. If $b$ has 0 prime factors, then $b=1$, and whatever the value of $a$ is, we have $(a, b)=1$, which is not a prime number.

b. If $b$ has exactly one prime factor, that is, it is itself prime, then the condition that $(a, b)$ and $(b, c)$ be prime means that both must coincide with $b$. The integers $a, c$ can therefore be chosen in any way among the divisors of $30=2 \cdot 3 \cdot 5$ that are multiples of $b$. It is easy to check that (whatever the value of $b$) there are 4 choices for $a$ and 4 choices for $c$ (for example, if $b=2$, then $a$ and $c$ can be chosen in the set $\{2,6,10,30\}$). Taking into account that $30=2 \cdot 3 \cdot 5$ has exactly 3 distinct prime factors (and therefore there are 3 choices for $b$ in this case), we have $3 \cdot 4 \cdot 4=48$ possible choices for the triple $(a, b, c)$.

c. If $b$ has exactly 2 prime factors, say $b=pq$, then setting $r=30 / b$ one easily sees that the condition that $(a, b)$ and $(b, c)$ be prime numbers is equivalent to the integers $a, c$ belonging to the set $\{p, q, pr, qr\}$. We observe that choosing $b$ is equivalent to choosing the prime $r \in \{2,3,5\}$, so in this case we have 3 choices for $b$ and 4 choices for each of $a$ and $c$, for a total of $3 \cdot 4 \cdot 4=48$ possible choices.

d. Finally, if $b$ has exactly 3 prime factors, then $b=30$. Since by hypothesis $a$ and $c$ divide $30$, we have $(a, b)=a$ and $(b, c)=c$, so we must choose both $a$ and $c$ in the set $\{2,3,5\}$. We then have $3 \cdot 3=9$ choices for the triple $(a, b, c)=(a, 30, c)$.

In total, the number of triples $(a, b, c)$ satisfying the desired properties is $48+48+9=105$.