Use euler's theorem to find a number x between 0 and 28 with x85 congruent to 6 modulo 35.

Problem: find 0 ≤ x ≤ 28 such that x^85 ≡ 6 modulo 35.

By Fermat-Euler theorem:
If a and n are coprime, i.e. (a,n)=1, then
a^phi(n) ≡ 1 mod n  
where phi(n)=totient function, the number of positive integers less than n that is coprime with n.

for n=35, phi(35)=24 calculated as follows:
There are 10 positive integers from 1 to 34 which are NOT coprime with 35, namely {5,7,10,14,15,20,21,25,28,30}.Therefore phi(35)=34-10=24

From Fermat-Euler theorem, 
x^(phi(35) = x^24 ≡ 1 modulo 35  since (24,35)=1, i.e. 24 and 35 are coprime.
=>
x^12 ≡ ± 1 modulo 35. (1)

and 

x^85 ≡ x^(85-3*24) ≡ x^(85-72) ≡ x^(13) ≡ 6 mod 35 (2)

Substituting (1) in (2)
x^(12)*x ≡ 6 mod 35
=>
(+1)*x = 6 mod 35   or  (-1)*x ≡ 6 mod 35
x ≡ 6 mod 35           x ≡ -6 mod 35   (rejected)

=>  x=6 

So

6^85 ≡ 6 mod 35

If any clarifications are needed or if you find any errors, please post.

It is going to be the third choice