**Option (C)** is correct. Reason is as follows :

Here, at first level we are Given $n$ elements, out of which we have to find smallest $3$ numbers.

We compare $2 -2$ elements as shown in figure & get $n/2$ elements at Second level.

**Note: Minimum element is in these **$n/2$ **elements.**

So, comparisons for this is $n/2$.

Similarly for next level we have $n/4$ Comparisons & $n/2$ elements and so on.

Total Comparisons till now is $n/2 + n/4 + n/8 + \ldots + 4 + 2 +1 = (2^{\log n}-1) = n-1$ {Use G.P. sum}

Now we have to get smallest $3$.

We have 1st smallest at last level already =>$0$ Comparison for this.

=> $2$nd & $3$rd smallest can be found in $O(\log n)$ time as shown below:

Minimum Element must have descended down from some path from top to Bottom.

=> SQUARES represent Candidates for $2$$^{nd}$ minimum.

Every element that is just below $m$1(first minimum) is a candidate for second minimum.

So, $O(\log n)$ Comparisons for finding second smallest.

=> Similarly for $3$$^{rd}$ minimum we get $O(\log n)$ time.. As, every element that is just below 1st & 2nd minimum is a candidate for $3$$^{rd}$ minimum.