10 V $\mathrm{H}_{2} \mathrm{O}_{2}$ means $1 \mathrm{L}$ of this solution will produce $10 \mathrm{LO}_{2}$ at STP.
$$
\begin{array}{c}
2 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\quad \mathrm{O}_{2} \\
68 \mathrm{g} \quad \quad \quad \quad 22.4 \text{L at STP}
\end{array}
$$
$\therefore 22.4 \mathrm{L}$ of $\mathrm{O}_{2}$ is obtained from $\mathrm{H}_{2} \mathrm{O}_{2}=68 \mathrm{g}$
10 L of $O_2$ will be obtained from
$$
\mathrm{H}_{2} \mathrm{O}_{2}=\frac{68}{22.4} \times 10=30.36 \mathrm{g}
$$
$1000 \mathrm{mL}$ of the given solution contains $30.36 \mathrm{g}$ $\mathrm{H}_{2} \mathrm{O}_{2}$ and $100 \mathrm{mL}$ of the given solution contains
$$
\frac{30.36 \times 100}{1000}=3.03 \mathrm{g} \mathrm{H}_{2} \mathrm{O}_{2}
$$
Thus, $\%$ strength of $\mathrm{H}_{2} \mathrm{O}_{2}$ is $3.03(\approx 3)$