Find the initial volume of the weak acid given a titration graph
Outline

There are three main points of interest in the titration graphic, and two you have found yourself. The three points are


The start at $V_\text{titr}=0$
Halfway to the equivalence point at $V_\text{titr}=10~\mathrm{mL}$
The equivalence point at $V_\text{titr} = 20~\mathrm{mL}$


Why is number 2 important? Because that is where we can read out the $\text{p}K_\mathrm{a}$ value of the acid, because at that point $[\ce{HA}] = [\ce{A-}]$.

Now, armed with this information, here is how to proceed:


Find out total amount of acid
Find out total concentration of acid
Calculate the initial volume from the data gathered in steps 1 and 2


1  Total amount of acid

The total amount of acid can be gleaned from information point 3, the equivalence point.

$$ n_\text{eq} = V_\text{titr,eq} \cdot c_\text{titr} = 20~\mathrm{mL} \cdot 0.150~\mathrm{mol\,L^{-1}} = 0.003~\mathrm{mol} $$

2  Total concentration of acid

The total concentration can be calculated from information points 1 and 2. We know the initial concentrations (denoted by subscript $_0$) because of the pH value at the beginning:
$$ [\ce{A-}]_0 = [\ce{H+}]_0 = 10^{-3}$$

Using the equilibrium constant that we got at point two, we can calculate the initial concentration of the associated (un-dissociated) acid species:
$$ [\ce{HA}]_0 = \frac{[\ce{H+}][\ce{A-}]}{K_\mathrm{a}} = \frac{10^{-6}}{10^{-5}} = 10^{-1} $$

The total initial concentration of all acid is then trivial:
$$ [\ce{A}]_\text{tot,0} = [\ce{HA}]_0 + [\ce{A-}]_0 = 0.101 $$

Note that we here actually calculated the activities and not concentrations. Since we assume an activity coefficient of $\gamma = 1$ because we only consider dissolved species, the concentrations are simply the activity multiplied with the standard concentration:
$$ c(\ce{A})_\text{tot,0} = \gamma \cdot [\ce{A}]_\text{tot,0} \cdot 1~\mathrm{mol\,L^{-1}} = 0.101~\mathrm{mol\,L^{-1}} $$

3  Calculation of the initial volume

The initial volume is now calculated following the simple concentration formula:
$$ c_0 = \frac{n_0}{V_0} \Leftrightarrow V_0 = \frac{n_0}{c_0}$$

Since the amount of acid is constant ($n = n_0\; \mathrm{const}$), we can easily fill in the values:
$$ V_0 = \frac{0.003~\mathrm{mol}}{0.101~\mathrm{mol\, L^{-1}}} \approx 0.0297~\mathrm{L} = 29.7~\mathrm{mL} $$