The smallest element in sample $S$ would be $X_i$ for which $i$ is smallest.
The given probability is for selection of each item of $X$. Independent selection means each item is selected with probability $\frac{1}{2}$.
Probability for $X_1$ to be smallest in $S = \frac{1}{2} $.
Value of $X_1=2$.
Probability for $X_2$ to be smallest in $S$ = Probability of $X_1$ not being in S $\times$ Probability of $X_2$ being in $S$ $= \frac{1}{2} . \frac{1}{2} $.
Value of $X_2=2^2=4$.
Similarly, Probability for $X_i$ to be smallest in $S = (1/2)^i$.
Value of $X_i=2^i$ .
Now Required Expectation= $\sum_{i=1}^{n}2^{^{i}} \times \left ( \frac{1}{2} \right )^{i} = \sum_{i=1}^n 1 = n $.
The answer is option D.