Chemistry ?

You can determine which form an indicator will be in by comparing the pH to the pKa of the indicator. For methyl orange, pKa = 4.0. The pKa is the pH at which half of the indicator is in the acidic form (HIN) and half is in the basic form (IN-). At a pH = pKa (pH 4 for this indicator) the solution would be orange (half red and half yellow). At pHs below the pKa, more and more is in the acidic form so the solution will be redder. At pH values above the pKa, more and more is in the basic form and so the solution will be more yellow.

Since pH = 6 is quite a bit above the pKa of the indicator, the solution will be yellow. 

HIn(aq) + H₂O(l) ⇌ In⁻(aq) + H₃O⁺(aq)

Henderson-Hasselbach equation:
pH = pKa + log([In⁻]/[HIn])
6 = -log(1 × 10⁻⁴) + log([In⁻]/[HIn])
6 = 4 + log([In⁻]/[HIn])
log([In⁻]/[HIn]) = 2
[In⁻]/[HIn] = 10²
[In⁻]/[HIn] = 100

Since [In⁻] ≫ [HIn], the solution shows the color of In⁻, i.e. yellow.