A set with $n$ elements has $2n$ subsets. Now, from the given set we can have
$\binom{n}{n}$ subsets of size $n$ (which will have $2^n$ subsets and a subset of subset will be a subset of a set)
$\binom{n}{n-1}$ subsets of size $n-1$ (which will have $2^{n-1}$ subsets which are also subsets of original set)
$\binom{n}{n-2}$ subsets of size $n-2$ (which will have $2^{n-2}$ subsets which are also subsets of original set)
$\cdots$
$\binom{n}{1}$ subsets of size $1$
$\binom{n}{0}$ subsets of size $0$.
So, total number of ordered pairs satisfying the subset order will be
$ 2^n \binom{n}{n} + 2^{n-1} \binom{n}{n-1} + 2^{n-2} \binom{n}{n-2} + \cdots + 2^0\binom{n}{0}$
$= (x + 1)^n\ where\ x = 2$,
$= 3^n$
