$x_1, x_2, x_3,…,x_n$ are integers, suppose $x_i$ is the largest among these integers, the number of bits it’ll contain will be $\log_2(x_i)$ bits, then the total number of bits for all integers will be $O(n\log_2(x_i))$, now in the worst case all of these bits will be input to 2-input AND/OR gate in the first level, so

$\text{number of gates in the first level = } O(n\log_2(x_i)), \\\text{number of gates in the second level} = \frac{O(n\log_2(x_i))}{2}, \\\text{number of gates in the thirdd level} = \frac{O(n\log_2(x_i))}{2^2},\\…\\\text{number of gates in the final level} = 1 (\text{as mapping is from integers to 0/1})$

$$\text{Total number of gates in all levels} = \log_2(x_i)[n + \frac{n}{2} + \frac{n}{2^2} + \frac{n}{2^3} + … 1] \\= \log_2(x_i)\times(2^{\log_2(n)}-1)$$

Now here, both a and c are possible answers depending on if we consider the value of $x_i$ similar to the constant in non-comparison sorts.

$\text{number of gates in the first level = } O(n\log_2(x_i)), \\\text{number of gates in the second level} = \frac{O(n\log_2(x_i))}{2}, \\\text{number of gates in the thirdd level} = \frac{O(n\log_2(x_i))}{2^2},\\…\\\text{number of gates in the final level} = 1 (\text{as mapping is from integers to 0/1})$

$$\text{Total number of gates in all levels} = \log_2(x_i)[n + \frac{n}{2} + \frac{n}{2^2} + \frac{n}{2^3} + … 1] \\= \log_2(x_i)\times(2^{\log_2(n)}-1)$$

Now here, both a and c are possible answers depending on if we consider the value of $x_i$ similar to the constant in non-comparison sorts.