10 million $=10^{-7}$
$\therefore 10$ million $^{\text {th }}$ part of $1.33 cm ^{3}$
$=1.33 \times 10^{-7} cm ^{3}$
For pure water, $=1.33 \times 10^{-7} mL \\\left[ H ^{+}\right] =10^{-7} mol / L $
$\Rightarrow 1 L$ water contains $\left.H ^{+}\right]=10^{-7} mol$
or $1 mL$ water contains $\left[H^{+}\right]=\frac{10^{-7}}{1000}=10^{-10} mol$
or 10 million $^{\text {th }}$ part of $1 \cdot 33 cm ^{3}$ water contains
$\left[ H ^{+}\right] =1 \cdot 33 \times 10^{-7} \times 10^{-10} mol$
$=1 \cdot 33 \times 10^{-17} mol$
$\therefore $ Number of $H ^{+}$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