Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11}g,$ where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is 2/3 times that of the earth. If the  escape speed on the surface of the earth is taken to be 11km/s, the escape velocity on the surface of the planet (in km/s) will be …….

The acceleration due to gravity on the surface of a planet of mass $M_p$ and radius $R_p$ is given by
$9_p=G{M}_p/R_p^2.$    (1)
The mass is related to density ${\rho }_p$ by  $M_p=(4/3)\pi R_p^3{\rho }_p.$
Substitute in equation (1) and simplify to get
  $R_p=3g_p/\left(4\pi G{\rho }_p\right).$
The escape velocity from the surface of the planet is given by
  ${\upsilon }_p=\sqrt{2G{M}_p/R_p}=\sqrt{2g_p{R}_p}$
$=\sqrt{3g_p^2/\left(2\pi G{\rho }_p\right).}$          (2)
Substitute value in equation (2) to get
  $\frac{{\upsilon }_p}{{\upsilon }_e}=\sqrt{\frac{g_p^2/g_e^2}{{\rho }_p/{\rho }_e}}=\sqrt{\frac{6/121}{2/3}}=\frac{3}{11},$
Which gives  ${\upsilon }_p=3\text{\hspace{0.17em}km/s.}$