20 volume of ${\mathrm{H}}_2{\mathrm{O}}_2$ solution has a strength of about 

20 volume ${\mathrm{H}}_2{\mathrm{O}}_2$ solution means that 1 L of ${\mathrm{H}}_2{\mathrm{O}}_2$ forms 20 L of oxygen. 

Decomposition of hydrogen peroxide is:

$2{\mathrm{H}}_2{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$

Above reaction states that 68 g of hydrogen peroxide forms 22.4 L or one mole of oxygen gas. Mass of hydrogen peroxide required to produce 20 L of oxygen gas is calculated as:

$\begin{array}{ccc}22.4 \mathrm{L} \mathrm{of} \mathrm{oxygen} \mathrm{gas}& =& 68 \mathrm{g} \mathrm{of} \mathrm{hydrogen} \mathrm{peroxide}\\ 20 \mathrm{L} \mathrm{of} \mathrm{oxygen} \mathrm{gas}& =& 20 \mathrm{L} \mathrm{of} \mathrm{oxygen} \mathrm{gas}\times \frac{68 \mathrm{g} \mathrm{of} \mathrm{hydrogen} \mathrm{peroxide}}{22.4 \mathrm{L} \mathrm{of} \mathrm{oxygen} \mathrm{gas}}\\ & =& 60.7 \mathrm{g} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2\end{array}$

One liter or 1000 mL (or 1000 g) has 60.7 g of hydrogen peroxide. Percentage strength of hydrogen peroxide in the given solution is calculated as:

$\begin{array}{ccc}\mathrm{Percentage} \mathrm{strength}& =& \frac{\mathrm{Mass} \mathrm{of} \mathrm{hydrogen} \mathrm{peroxide}}{\mathrm{total} \mathrm{mass} \mathrm{of} \mathrm{compound}}\times 1000\\ & =& \frac{60.7 \mathrm{g}}{1000 \mathrm{g}}\times 100\\ & =& 6\%\end{array}$