6. Arrange the integers that satisfy the condition "contain at least one digit 0 and are multiples of 4" in ascending order:
$$
20,40,60,80,100,104, \cdots \text {. }
$$

Then the 158th number in this sequence is ( ).
(A) 2000
(B) 2004
(C) 2008
(D) 2012
6.C.

Among positive integers, the characteristic of being a multiple of 4 is that the last digit is a multiple of 4, which includes 7 forms containing the number 30: 00, 04, 08, 20, 40, 60, 80, and 18 scenarios not containing the number 30.

Obviously, there are only 4 two-digit numbers that satisfy the condition; there are a total of $9 \times 7=63$ two-digit numbers that satisfy the condition; and there are $7 \times 10 + 18 \times 1 = 88$ three-digit numbers that satisfy the condition and have the first digit as 1.

Since $4 + 63 + 88 = 155$, the 155th number that satisfies the condition, in ascending order, is 1980.
The next numbers that satisfy the condition are $2000, 2004, 2008$.
Therefore, the 158th number in this sequence is 2008.