To find the greatest number by which the product of three consecutive even numbers would be exactly divisible, we need to consider the prime factors of each number.

Let`s consider three consecutive even numbers:

- First even number: 2n, where n is any integer

- Second even number: 2n+2, which can be simplified as 2(n+1)

- Third even number: 2n+4, which can be simplified as 2(n+2)

Now, let`s find the prime factors of each number:

- First even number (2n) has a prime factor of 2.

- Second even number (2(n+1)) also has a prime factor of 2.

- Third even number (2(n+2)) has prime factors of 2 and 2.

To find the greatest number that will divide the product of these three numbers, we need to find the maximum power of each prime factor that appears in the numbers. In this case, the maximum power of 2 is 2, as it appears twice in the third even number.

Therefore, the greatest number that the product of three consecutive even numbers would be exactly divisible by is 2^2, which is 4. Hence, the correct answer