$''10$ volume $H_2O_2''$ means $1 \,mL$ of its solution on decomposition at $NTP$, give $10 \,mL$ oxygen gas. Volume of oxygen formed from $100 \,mL$ of solution at $NTP = 1000\,mL$
$\begin{matrix}2H_{2}O_{2}&\ce{->}&2H_{2}O+&O_{2}\\ 2 \text{moles}&&&1 \text{mole}\\ 2\times34\,g&&&22400\,mL\end{matrix}$
$\because 22400\, mL\, O_2$ formed at $NTP$ by decomposition of $68\,g \,H_2O_2$.
$\therefore 1\,mL\,O_{2}$ formed at $NTP$ from $\frac{68}{22400}$ of $H_{2}O_{2}$
$\therefore 1000\,mL\,O_{2}$ formed at $NTP$ from
$\frac{68\times1000}{22400}g\,H_{2}O_{2}=3.0
35\,g\,H_{2}O_{2}$
So, concentration of $''10$ volume $H_2O_2’’$
$=3.0\%$ approximately