Which of the following cannot be the units digit of a perfect square: 6, 1, 9, 8?

The units digit of a perfect square is determined by the units digit of the number being squared.
Let's list the units digits of squares of single-digit numbers:

• $0^2 = 0$
• $1^2 = 1$
• $2^2 = 4$
• $3^2 = 9$
• $4^2 = 16 \implies 6$
• $5^2 = 25 \implies 5$
• $6^2 = 36 \implies 6$
• $7^2 = 49 \implies 9$
• $8^2 = 64 \implies 4$
• $9^2 = 81 \implies 1$

The possible units digits of a perfect square are 0, 1, 4, 5, 6, and 9.

• (a) 6: This can be a units digit of a perfect square (e.g., $4^2 = 16$).
• (b) 1: This can be a units digit of a perfect square (e.g., $1^2 = 1$).
• (c) 9: This can be a units digit of a perfect square (e.g., $3^2 = 9$).
• (d) 8: This is not in the list of possible units digits (0, 1, 4, 5, 6, 9).

Therefore, 8 cannot be the units digit of a perfect square number.

The correct option is (d).

$$\boxed{\text{(d) 8}}$$