Time for the return journey
This transfer orbit is geometrically equivalent to the transfer orbit for the forward journey, for both of them are heliocentric and share the same aphelion and perihelion. The time for the transfer is thus the same as the time needed for the initial journey to Mars :
Ttransfer = T/2 = pi*[(dEarth+dMars)^3/(8*G*Ms)]
That is, again Ttransfer=22,366,601 seconds which are equivalent to approximately 259 days or eight and a half months.