6. The first digit is 1, and there are exactly two identical digits in the four-digit number ( ).
(A) 216
(B) 252
(C) 324
(D) 432
6. (D).

The qualified four-digit numbers must contain one 1 or two 1s.
(1) Contain two 1s. Choose two digits from the remaining 9 digits other than 1, which has $C_{9}^{2}$ ways, and then form any permutation of a three-digit number with one of the 1s, which has $\mathrm{P}_{3}^{3}$ ways. Thus, the four-digit numbers with the first digit being 1 are $N_{1}=C_{9}^{2} \cdot \mathrm{P}_{3}^{3}$ in total.
(2) Contain only one 1. Choose two digits from the remaining 9 digits, which has $\mathrm{C}_{9}^{2}$ ways, and one of these digits is repeated, which has $\mathrm{C}_{2}^{1}$ ways, and the three digits form a three-digit number, which has $\mathrm{P}_{3}^{1}$ ways. Therefore, the four-digit numbers with the first digit being 1 are $N_{2}=\mathrm{C}_{9}^{2} \mathrm{C}_{2}^{1} \mathrm{P}_{3}^{1}$ in total.
Hence, the total number of four-digit numbers that meet the criteria is
$$
N=N_{1}+N_{2}=432 \text { (numbers). }
$$