Let $n_l$ and $n_r$ nodes be present in the left and right sub-trees respectively.
We have, $\frac{n_r}{2}\le n_l \le 2n_r$. Without loss of generality, let the left sub-tree have greater number of nodes $(2n_r\ nodes)$. Then, $n_r + 2n_r + 1 = n$. Thus we get, $n_l = {2(n-1) \over 3}$ and $n_r = {n-1 \over 3}$.
Total number of nodes can be described by the recurrence:
$T(n) = T\left(\frac{n-1}{3}\right) + T\left(\frac{2(n-1)}{3}\right) + 1 \text{ and } T(1) = 1$
As this makes maximum nodes go to one subtree and that is what we want to get the maximum height with a given number of nodes.
Now, the height of the tree will be: $H(n) = H(\frac{2}{3}(n-1)) + 1$ and $H(1) = 1$
We can draw a recurrence tree and the cost at each level is 1, and the height will be $\color{red}{\log_{\frac{3}{2}}(n)}$.
So, D option is the answer.