10 million $=10^{-7}$
$\therefore$ 10 $million^{th}$ part of $1.33 \mathrm{cm}^{3}=1.33 \times 10^{-7} \mathrm{cm}^{3}$ $=1.33 \times 10^{-7} \mathrm{mL}$
For pure water,
$$
\begin{aligned}
\left[\mathrm{H}^{+}\right] &=10^{-7} \mathrm{mol} / \mathrm{L}
\end{aligned}
$$
$\Rightarrow 1$ L water contains $\left[\mathrm{H}^{+}\right]=10^{-7} \mathrm{mol}$
or $1 \mathrm{mL}$ water contains $\left[\mathrm{H}^{+}\right]=\frac{10^{-7}}{1000}=10^{-10} \mathrm{mol}$
or 10 million $^{\text {th }}$ part of $1.33 \mathrm{cm}^{3}$ water contains $\begin{aligned}\left(\mathrm{H}^{+}\right] &=1.33 \times 10^{-7} \times 10^{-10} \mathrm{mol} \\ &=1.33 \times 10^{-17} \mathrm{mol} \end{aligned}$
$\therefore$ Number of $\mathrm{H}^{+}$ ions $=1.33 \times 10^{-17} \times N_{A}$
$=1.33 \times 10^{-17} \times 6.022 \times 10^{23}$
$=8.009 \times 10^{6}=8.01$ million