1. Use the relation between surface gravity, density, and radius

For a spherical planet, $g = \frac{GM}{R^2}$ and since $M = \frac{4}{3}\pi R^3 \rho,$ we get $g = \frac{G\left(\frac{4}{3}\pi R^3\rho\right)}{R^2} = \frac{4}{3}\pi G \rho R.$

So, $g \propto \rho R.$

2. Compare planet with earth

Let the planet quantities be subscript $p$ and earth quantities be subscript $e$.

Given: $g_p = \frac{\sqrt{6}}{11} g_e$ $\rho_p = \frac{2}{3}\rho_e$

Using $\frac{g_p}{g_e} = \frac{\rho_p R_p}{\rho_e R_e},$ we get $\frac{\sqrt{6}}{11} = \frac{2}{3}\cdot \frac{R_p}{R_e}.$

Hence, $\frac{R_p}{R_e} = \frac{\sqrt{6}}{11}\cdot \frac{3}{2} = \frac{3\sqrt{6}}{22}.$

3. Use escape speed formula

Escape speed is $v_e = \sqrt{2gR}.$

Therefore, $\frac{v_p}{v_e} = \sqrt{\frac{g_p R_p}{g_e R_e}}.$

Substitute the ratios: $\frac{v_p}{v_e} = \sqrt{\frac{\sqrt{6}}{11} \cdot \frac{3\sqrt{6}}{22}}.$

Now, $\sqrt{6}\cdot \sqrt{6} = 6,$ so $\frac{v_p}{v_e} = \sqrt{\frac{18}{242}} = \sqrt{\frac{9}{121}} = \frac{3}{11}.$

4. Find escape speed on the planet

Given earth's escape speed: $v_e = 11\ \text{km s}^{-1}$

So, $v_p = \frac{3}{11}\times 11 = 3\ \text{km s}^{-1}.$

5. Final answer

The escape speed on the planet is $\boxed{3}$ (in km s$^{-1}$).