Escape velocity of a body from earth is 11.2 km s$$^{-1}$$. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is:

Escape velocity: $$v_e = \sqrt{\frac{2GM}{R}}$$.

Given: $$R_p = \frac{R_e}{3}$$ and $$M_p = \frac{M_e}{6}$$.

$$\frac{v_p}{v_e} = \sqrt{\frac{M_p/R_p}{M_e/R_e}} = \sqrt{\frac{M_p \cdot R_e}{M_e \cdot R_p}} = \sqrt{\frac{(M_e/6) \cdot R_e}{M_e \cdot (R_e/3)}} = \sqrt{\frac{1/6}{1/3}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$

$$v_p = \frac{11.2}{\sqrt{2}} = \frac{11.2 \times \sqrt{2}}{2} = 5.6\sqrt{2} \approx 7.92 \text{ km/s}$$

The answer is Option (4): $$\boxed{7.9 \text{ km s}^{-1}}$$.