The worst case running time to search for an element in a balanced binary search tree with $n2^n$ elements is

In a balanced binary search tree with $N$ elements, the height of the tree is $\Theta(\log N)$.

The worst-case search time is proportional to the height of the tree:

Given $N = n \cdot 2^n$:

As $n \to \infty$, the linear term $n$ dominates the logarithmic term $\log n$. Therefore:

Hence, the correct option is (C).