If earth has a mass nine times and radius twice to the of a planet...

The formula to calculate the escape velocity$\left({v}_{e}\right)$ for the Earth is given by

${v}_{e} = \sqrt{\dfrac{2G{M}_{e}}{{R}_{e}}} . . . . . . . . . . . . . . . . . . . . . . . ( 1 )$

where, $G$ is the Universal Gravitational constant, ${M}_{e}$ is the mass and ${R}_{e}$ is the radius of Earth.

The formula for the escape velocity $\left({v}_{esc}\right)$ of any planet is given by

${v}_{esc} = \sqrt{\dfrac{2GM}{R}} . . . . . . . . . . . . . . . . . . . . . . . . ( 2 )$

where, $M$ is the mass and $R$ is the radius of the planet.

Given that the mass of the planet is $9$ times the mass of the Earth and the radius of the planet is $2$ times the radius of the Earth.

Substitute $\dfrac{{M}_{e}}{9}$ for $M and \dfrac{{R}_{e}}{2}$ for $R$ into equation (2) and simplify to obtain the escape velocity for the planet in terms of that in Earth.

$\begin{cases}{v}_{esc} & = & \sqrt{\dfrac{2G\dfrac{{M}_{e}}{9}}{\dfrac{{R}_{e}}{2}}} \\ = & \sqrt{\dfrac{2}{9}\dfrac{2G{M}_{e}}{{R}_{e}}} \\ = & \dfrac{\sqrt{2}}{3}{v}_{e} . . . . . . . . . . . . . . . . . . . . . ( 3 )\end{cases}$

Comparing equation (3) with the given expression, it can be written that

$x = 2$.