How to calculate the sufficient ratio of acid and corresponding base of the indicator that allows to clearly observe the endpoint of a titration?
If you assume that the $pK_{a}$ of the indicator is 7.0 and you assume that you're going to the end point at pH 8.5 then the Henderson Hasselbach equation


  
    ${pH = pK_{a} + log \frac{[In^-]}{[HIn]}}$
  


can be rearranged to:


  
    $\frac{[HIn]}{[In^-]} = 10^{(pH - pK_{In})} = 31.62+ \approx 32$
  


However:

(1) There is no real statement in the problem that $pK_{a}$ of the indicator is 7.0 which was assumed from:


  
    $\dfrac{1}{2}(5.5 + 8.5) = 7.0$
  


There is little real chemical validity behind this assumption. 

(2) There isn't a real justification of why color at pH 8.5 is any better than color at pH 5.5. So the ratio is either 32 or 1/32. Flip a coin :-(