2. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$ and $a_{1}=9$, with the sum of its first $n$ terms being $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is
A. 5
B. 6
C. 7
D. 8
2. C From $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right), a_{1}-1=8$ we know that the sequence $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$, so $S_{n}-n=\frac{8 \times\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{1+\frac{1}{3}}=6-6 \times\left(-\frac{1}{3}\right)^{n}$,
$$
\left|S_{n}-n-6\right|=6 \times\left(\frac{1}{3}\right)^{n}250>3^{5}$, solving it yields $n-1>5$, i.e., $n>6$. Thus, $n \geqslant 7$. It is not difficult to verify that when $n=7$, it meets the requirements of the problem. Therefore, the answer is $\mathrm{C}$.