The answer to this says that the resulting normal form is $2\times 4\times 4$
This is what I've tried so far:
For the subgame of player $2$ and $3$ where player $2$ takes dilemma (2 is row, 3 is column): \begin{array}{c|lcr} & FF & FM & MF & MM \\ \hline F & 2,2 & 2,2 & 5,1 & 5,1 \\ M & 1,5 & 4,4 & 1,5 & 4,4 \end{array}
For the subgame of player $2$ and $3$ where player $2$ takes sexes (2 is row, 3 is column):
\begin{array}{c|lcr} & MM & MB & BM & BB \\ \hline M & 3,2 & 3,2 & 1,1 & 1,1 \\ B & 0,0 & 2,3 & 0,0 & 2,3 \end{array}
For the subgame between $1$ and $2$, here is the normal form, which I'm not quite sure how to construct:
\begin{array}{c|lcr} & DF & DM & SM & SB \\ \hline D & 2,3 & 3,1 & - & - \\ S & - & - & 0,3 & 3,2 \end{array}
I'm not quite sure what goes into the blanks. Also not quite sure what would be the last $2\times 4$ normal form game. Any help is appreciated!
Note the dotted line between the results from 2’s actions: this disqualifies FM, MF, BM, MB as valid actions because you can’t plausibly follow through. Now you can see the valid actions are:
You should be able to construct the 2x4x4 matrix by following the results of these actions.