Solution:

Division by 70 leaves a remainder of 0 if and only if the number is divisible by both 7 and 10.

We know that 10 = 2 x 5 and 70 = 2 x 5 x 7. Therefore, we need to check if x is divisible by 2, 5, and 7.

Checking for Divisibility by 2:

The last digit of each of the four numbers in x is odd, so their product is odd, and therefore, not divisible by 2.

Checking for Divisibility by 5:

The last digit of 16^3 is 6, the last digit of 17^3 is 3, the last digit of 18^3 is 2, and the last digit of 19^3 is 9. Therefore, their product ends with the digit 6 x 3 x 2 x 9 = 324, which means it is divisible by 5.

Checking for Divisibility by 7:

We can use the divisibility rule for 7, which states that if we subtract twice the units digit from the remaining digits of a number, and the result is divisible by 7, then the number is also divisible by 7.

For example, applying this rule to 476, we get 47 - 2 x 6 = 35, which is divisible by 7, so 476 is also divisible by 7.

Using this rule, we get:

16^3 ≡ 1 - 2 x 6 ≡ -11 ≡ 4 (mod 7)

17^3 ≡ -1 (mod 7)

18^3 ≡ 1 - 2 x 8 ≡ -15 ≡ 1 (mod 7)

19^3 ≡ -1 (mod 7)

Therefore, their product is:

x ≡ 4 x (-1) x 1 x (-1) ≡ 4 (mod 7)

Therefore, x is divisible by 5 but not by 2 or 7.

The only option among the given choices is 69.

Therefore, the answer is (c) 69.