To find the maximum value of the cross product of two vectors, we need to determine the maximum value of the magnitude of the cross product.

Let's assume the two vectors are A and B. The magnitude of the cross product of two vectors is given by the formula:

|A x B| = |A| * |B| * sin(theta)

where |A| and |B| are the magnitudes of vectors A and B, and theta is the angle between the two vectors.

We are given that the maximum value of the vector sum of A and B is 6, which means that the magnitude of the vector sum |A + B| is 6.

Now, let's consider the scenario where the vector sum |A + B| is maximum.

- The maximum value of the vector sum occurs when the vectors A and B are aligned in the same direction, resulting in constructive interference.
- In this case, the magnitude of the vector sum |A + B| is equal to the sum of the magnitudes of A and B, i.e., |A| + |B| = 6.

Using this information, we can deduce the following:

|A| + |B| = 6

Since the magnitude of a vector cannot be negative, we can rewrite the above equation as:

|A| = 6 - |B|

- The magnitude of the cross product is maximum when the angle between the vectors is 90 degrees, resulting in the maximum value of sin(theta) which is 1.
- In this case, the magnitude of the cross product |A x B| is given by |A| * |B| * sin(theta) = |A| * |B|.

Using the equation |A| + |B| = 6, we can substitute |A| = 6 - |B| into the equation for the magnitude of the cross product:

|A x B| = (6 - |B|) * |B|

To find the maximum value of |A x B|, we need to find the maximum value of the expression (6 - |B|) * |B|.

- We can rewrite the expression as a quadratic equation: f(|B|) = -|B|^2 + 6|B|.
- The maximum value of a quadratic equation occurs at the vertex, which is given by -b/2a, where a = -1 and b = 6.
- Plugging in the values, we get |B| = -6/(-2) = 3.

Therefore, the maximum value of the magnitude of the cross product |A x B| is |A| * |B| = (6 - 3) * 3 = 9.

Hence, the correct option is B) 16.