contribution of a worker =number of days he/she worked$*$number of hours worked per day$*$work done per hour.
Now coming to problem,
Let S,E,F does work of x units/hour.Since M is twice as efficient as others,M does work of 2x units/hour.
Let E works for y days.Then M works for y$/$2 days.
M works for 6 hrs in a day while E works for 12 hrs in a day.
$\therefore$ Contribution of M = (y$/$2)$*$2x$*$6 = 6xy
and Contribution of E = y$*$x$*$12$ = $12xy
Contribution of M$/$Contribution of E $=$1$/$2
The work done can be any thing.It can be eating 2 apples per hour or typing 3 pages per hour...depending on the project.