As illustrated in the figure, a water molecule has a specific...

Let the oxygen atom be at the origin $(0,0)$.

The y-axis is taken along the bisector of the H-O-H angle. The angle of each O-H bond with the y-axis is \theta = \dfrac{104^{}^{{}^{\circ}}}{2} = 52^{}^{{}^{\circ}}.

The coordinates of the two hydrogen atoms are (d \sin 52^{}^{{}^{\circ}}, -d \cos 52^{}^{{}^{\circ}}) and (-d \sin 52^{}^{{}^{\circ}}, -d \cos 52^{}^{{}^{\circ}}), where $d = 0.96 \times 10^{-10}\text{ m}$.

Mass of oxygen atom, $m_O = 16\text{ u}$
Mass of hydrogen atom, $m_H = 1\text{ u}$

By symmetry, the x-coordinate of the centre of mass is $0$.

The y-coordinate of the centre of mass is:
$y_{cm} = \dfrac{m_O y_O + m_H y_{H_1} + m_H y_{H_2}}{m_O + 2m_H}$

y_{cm} = \dfrac{16(0) + 1(-d \cos 52^{}^{{}^{\circ}}) + 1(-d \cos 52^{}^{{}^{\circ}})}{16 + 1 + 1}

y_{cm} = \dfrac{-2d \cos 52^{}^{{}^{\circ}}}{18} = -\dfrac{d \cos 52^{}^{{}^{\circ}}}{9}

The distance of the centre of mass from the oxygen atom is:
r = |y_{cm}| = \dfrac{d \cos 52^{}^{{}^{\circ}}}{9}

Substituting the given values (using \cos 52^{}^{{}^{\circ}} \approx 0.6156):
$r = \dfrac{0.96 \times 10^{-10} \times 0.6156}{9}$

$r \approx 0.0656 \times 10^{-10}\text{ m} = 6.56 \times 10^{-12}\text{ m}$

Rounding off, we get $r \approx 6.6 \times 10^{-12}\text{ m}$.

Answer: $6.6 \times 10^{-12}\text{ m}$