For any arrangement, we say that a position $k \in \{2, 3, \dots, 3n\}$ is a change if and only if the coin at position $k$ has a different colour to the coin at position $k-1$. Note that the number of blocks in any arrangement is one greater than the number of changes in that arrangement. For instance, the example arrangement given in the problem statement contains 5 changes (at positions 4, 6, 7, 9 and 10), and the number of blocks is $5+1=6$.

Next we count the total number of changes in all distinct coin arrangements, partitioning the count according to the position $k \in \{2, 3, \dots, 3n\}$ at which the change occurs (denote this total number of changes by $N_c$). If position $k$ is a change, we can choose the coin at position $k$ in 3 ways and the coin at position $k-1$ in 2 ways. For the remaining positions we can arrange the coins in $(3n-2)!/(n!(n-1)!(n-1)!)$ ways. Since the number of blocks is always one greater than the number of changes, we obtain
$$N_c = (3n - 1) \cdot 3 \cdot 2 \cdot \frac{(3n - 2)!}{n!(n - 1)!(n - 1)!} = 2n \cdot \frac{(3n)!}{(n!)^3}$$
where the factor $3n-1$ accounts for all of the possible change positions i.e., all possible values of $k \in \{2, 3, \dots, 3n\}$.

Since the number of blocks is always one greater than number of changes, the total number of blocks $N_b$ over all distinct coin arrangements, is obtained from $N_c$ by adding $(3n)!/(n!)^3$, which is equal to the total number of distinct coin arrangements, hence
$$N_b = N_c + \frac{(3n)!}{(n!)^3} = (2n + 1)\frac{(3n)!}{(n!)^3}$$
Thus the average number of blocks over all distinct coin arrangements is
$$N_b = N_c \cdot \left[ \frac{(3n)!}{(n!)^3} \right]^{-1} = 2n + 1.$$
Thus we have established the result, with $A = 2$ and $B = 1$.