Answer: 8/27

Explanation:  Given that w = 3k^3 + 2k^2 + k, we can find the group velocity as follows:

dw/dk = 9k^2 + 4k + 1

To find the point of maximum curvature, we need to set the second derivative of the dispersion relation equal to zero:

d^2w/dk^2 = 18k + 4 = 0

Solving for k, we get:

k = -2/9

Substituting this value of k into the expression for dw/dk, we get:

dw/dk = 9(-2/9)^2 + 4(-2/9) + 1

dw/dk = 1/3

Therefore, the group velocity of the wave is 1/3 times the phase velocity, which is given by the derivative of the dispersion relation with respect to frequency:

v_p = dw/dk = 9k^2 + 4k + 1

At the point of maximum curvature, we have:

v_p = dw/dk = 9(-2/9)^2 + 4(-2/9) + 1 = 8/9

Therefore, the group velocity of the wave is:

v_g = (1/3) v_p = (1/3)(8/9) = 8/27