Let U=$\{1,2,3\}$
All Possible subsets of U = $\{\phi,\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\}\}$
$A = {(x, X), x ∈ X \ and\ X ⊆ U}$
$x$ can be only $\{\phi,1,2,3\}${Considering $\phi$ also a valid $x$}
Now A can have elements like :- $\{1,\{1,2\}\}$ {Here X is $\{1,2\}$ }.
When $x=1$ , total elements in A = 4.
Similarly for $x=2$ and $x=3$ , total elements in A would be 4,4 respectively.
Therefore, total elements in A , $i.e \ |A|$ = $4+4+4 = 12$.
Not considering $x=\phi$ and $X=\phi$ since $\phi \in \phi $ is not true.
Now comparing against options ,
option (i) -> $n*2^{(n-1)}$ when n=3 , this option gives $12$ , matches.
option (ii) -> $\sum_{k=1}^{n}k*n_{C_{k}} = 1*3_{C_{1}} + 2*3_{C_{2}}+3*3_{C_{3}} = 3 + 6 +3 = 12.$
Thus both are correct.