6 If and only if $n=3 k-2(k=1,2, \cdots)$, then $3 \times a_{n}$. Therefore,
$$
b_{k}=\frac{(3 k-2)(3 k-1)}{2}=1+\frac{9 k(k-1)}{2},
$$

Thus,
$$
b_{200}=1+\frac{9 \times 200 \times 199}{2}=179101 .
$$

Hence, the answer is B.
6 Let $a_{n}=\frac{n(n+1)}{2}$, from the sequence $\left\{a_{n}\right\}$, remove all numbers that can be divided by 3, and the remaining terms are arranged in ascending order to form the sequence $\left\{b_{n}\right\}$, then $b_{200}=(\quad$.
(A) 200808
(B) 179101
(C) 153201
(D) 116808