Answer
$2\times(n!)^2$
Work Step by Step
We have 2 options as we can start the row by either men or women.
Let's say we start with a woman. Now because men and women alternate, women must occupy all the odd places and men must occupy all the even places. So women can shuffle among themselves(at odd places) in $n!$ ways and so can men. So we have $n! \times n!$ possibilities.
Similarly, if we start with a man, we have $n! \times n!$ possibilities.
So, in total, we have $n! \times n!+n! \times n!$ = $2\times(n!)^2$ total possible arrangements.