To solve for $a$, $b$, and $c$ to form the sides of a triangle, it is clear that they cannot be 0, i.e., $a, b, c \in \{1, 2, \cdots, 9\}$.

(1) If they form an equilateral triangle, let the number of such three-digit numbers be $n_{1}$. Since all three digits in the three-digit number are the same, we have:
$$
n_{1} = \mathrm{C}_{9}^{1} = 9.
$$

(2) If they form an isosceles (non-equilateral) triangle, let the number of such three-digit numbers be $n_{2}$. Since the three-digit number contains only 2 different digits, let them be $a$ and $b$. Noting that the legs and base of the triangle can be interchanged, the number of digit pairs $(a, b)$ is $2 \mathrm{C}_{9}^{2}$. However, when the larger number is the base, if $a > b$, it must satisfy $b < a < 2b$. In this case, the digit pairs that cannot form a triangle are as follows, totaling 20 cases:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline$a$ & 9 & 8 & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\
\hline$b$ & 4,3 & 4,3 & 3,2 & 3,2 & 2,1 & 2,1 & 1 & 1 & \\
\hline
\end{tabular}

At the same time, for each digit pair $(a, b)$, the two digits can be placed in three positions in $\mathrm{C}_{3}^{2}$ ways. Therefore,
$$
n_{2} = \mathrm{C}_{3}^{2} \left(2 \mathrm{C}_{9}^{2} - 20\right) = 6 \left(\mathrm{C}_{9}^{2} - 10\right) = 156.
$$

In summary, $n = n_{1} + n_{2} = 165$. Therefore, the answer is C.
(5) Let the three-digit number $n=\overline{a b c}$, if the lengths of the sides of a triangle can be formed by $a, b, c$ to constitute an isosceles (including equilateral) triangle, then the number of such three-digit numbers $n$ is ( ).
(A) 45
(B) 81
(C) 165
(D) 216