Physics light years. i showed my work. what did i do wrong

An astronaut travels to a distant star with a speed of 0.53c relative to Earth. From the astronaut's point of view, the star is 7.1 Light Years from Earth. On the return trip, the astronaut travels with a speed of 0.90 c relative to Earth.
What is the distance covered on the return trip, as measured by the astronaut? Give your answer in light-years.

I got 8.37 light years but that's wrong :(

heres my work:

t(initial)=D/v
t(initial)=7.1/.53
=13.40 yrs

t=t(initial)gamma
t=(13.40)(1-.53^2)^ (-1/2)
=15.80 yrs

d=vt
(15.80)(365)(24)(3600)(.53)(3x10^8)
= 7.92x106
= 8.37 light years

but thats wrong. where did i make a mistake?

The astronaut perceives the distance as 7.1 light years traveling at 0.53 c
7.1 = L*(√(1-0.53^2)) = L*0.848
L =8.37 light years, this would be the distance if the astronaut was at rest

Lr = 8.37*(√(1-0.90^2))= 8.37*0.436 = 3.65 light years
The astronaut perceives the return distance as 3.65 light years

You can do this whole problem in one step; you just need to know the length contraction equation:
L = L(o) / gamma
On the way there, the gamma factor is 1.179
On the way back, the gamma factor is 2.294
(7.1)/(1.179) + (7.1)/(2.294) = 9.12 light-years