A planet having mass $$9 M_e$$ and radius $$4R_e$$, where $$M_e$$ and $$R_e$$ are mass and radius of earth respectively, has escape velocity in km s$$^{-1}$$ given by: (Given escape velocity on earth $$V_e = 11.2 \times 10^3$$ m s$$^{-1}$$)

Given: Planet has mass $$M_p = 9M_e$$ and radius $$R_p = 4R_e$$. Escape velocity on Earth $$V_e = 11.2 \times 10^3$$ m/s.

The escape velocity formula is:

$$V_{esc} = \sqrt{\frac{2GM}{R}}$$

The ratio of escape velocities:

$$\frac{V_p}{V_e} = \sqrt{\frac{M_p}{M_e} \cdot \frac{R_e}{R_p}} = \sqrt{\frac{9M_e}{M_e} \cdot \frac{R_e}{4R_e}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$$

Therefore:

$$V_p = \frac{3}{2} \times V_e = \frac{3}{2} \times 11.2 = 16.8 \text{ km/s}$$

Therefore, the correct answer is Option B: $$\mathbf{16.8}$$.