How to determine the minimum pH at which iron(III) hydroxide dissolves completely?
The OP first stated the problem as:


  Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $\pu{10 mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.


I took that to mean that acid was added to a solution containing 10 mg of iron (iii) hydroxide and the pH of the solution when the iron (iii) hydroxide was totally dissolved was desired.*

Knowing that $K_\mathrm{sp} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ then
\begin{align}
  K_\mathrm{sp} &amp;= \ce{[Fe^{3+}][OH^-]^3} = 6\cdot10^{-38},&amp;
\text{ so }&amp;&amp;
\ce{[OH^-]} &amp;=\sqrt[3]{\frac{6\cdot10^{-38}}{\ce{[Fe^{3+}]}}}
\end{align}

You need to solve for $\ce{[Fe^{3+}]}$ which you can calculating knowing that there is $\pu{10 mg}$ of $\ce{Fe(OH)_3}$ in 0.1 liters of solution. It has a molecular weight of $\pu{107 g/mol}$, so there is 
$$0.010/107 = \pu{9.35* 10^-5 mol//L}\text{ of }\ce{Fe(OH)_3}.$$ 

Going back to the equation for the $K_\mathrm{sp}$:
$$\ce{[OH^-]} =\sqrt[3]{\frac{6\cdot10^{-38}}{\ce{[Fe^{3+}]}}} = \sqrt[3]{\frac{6\cdot10^{-38}}{9.35\cdot10^{-5}}} = 8.63\cdot10^{-12}$$

Knowing $\ce{[OH^-]}$ you can calculate $\ce{[H^+]}$ via

$$\ce{[H^+]} = \frac{1\cdot10^{-14}}{\ce{[OH^-]}} = \frac{1\cdot10^{-14}}{8.63\cdot10^{-12}} = 1.16\cdot10^{-3}$$

and then pH: 

$$\mathrm{pH} = -\log\left(1.16\cdot10^{-3}\right) = 2.94.$$



Now the OP has modified the problem statement to be:


  Knowing $K_\mathrm{s} = 6\cdot10^{-38}$ for $\ce{Fe(OH)3}$ in neutral solutions, calculate the minimum pH of an acidic solution in order to completely dissolve $10\ \mathrm{mg}$ of $\ce{Fe(OH)3}$. Data: $V = \pu{0.1 L}$.


We again end up with the final pH being just acidic enough to dissolve $\pu{10 mg}$ of iron (iii) hydroxide in $\pu{100 ml}$ of water, but obviously the solution must start out more acidic. The final solution will again have a pH of $2.94$ or a $\ce{[H^+]} = 1.16\cdot10^{-3}$.

In $\pu{100 mL}$ of a solution of pH $2.94$ there are $\pu{0.116 mmol}$ of acid. 

We calculated before that there were $\pu{9.35*10^{-5} mol}$ of $\ce{Fe^{3+}}$, but there are three $\ce{OH^-}$ anions for every cation of $\ce{Fe^{3+}}$, so there are $3 \times 9.35\cdot10^{-5} = \pu{2.81*10^{-4} mol}$ of $\ce{OH^-}$. 

The total amount of substance (in millimoles) of acid needed is 
$$0.116 + 0.281 = 0.397.$$

For $\pu{100 ml}$ to contain $\pu{0.397 mmol}$ the solution would need to be 3.97 millimolar in $\ce{[H^+]}$, or the pH = $2.40$. 

If you start with $\pu{100 ml}$ of solution with pH $2.40$ and dump in $\pu{10 mg}$ of $\ce{Fe(OH)3}$, then the final pH will be $2.94$ and all of the $\ce{Fe(OH)3}$ will be just dissolved.