C

Step 1: Understand the forces acting on the object when thrown upward. The gravitational force ( ext{Weight}) is given by $F_g = mg = 5 ext{ kg} \times 10 ext{ ms}^{-2} = 50 ext{ N}$.
Step 2: The retarding force (air resistance) is given as $F_r = 10 ext{ N}$. When thrown upward, the net force $ ext{F}_{net ext{up}}$ will be $F_g + F_r = 50 ext{ N} + 10 ext{ N} = 60 ext{ N}$.
Step 3: Using Newton's second law, the acceleration during ascent is $a_{up} = \frac{F_{net}}{m} = \frac{60 ext{ N}}{5 ext{ kg}} = 12 ext{ ms}^{-2}$ (downward).
Step 4: During descent, the total force acting on the object becomes $F_g - F_r = 50 ext{ N} - 10 ext{ N} = 40 ext{ N}$. Here, the net force during descent is $F_{net ext{down}} = 40 ext{ N}$.
Step 5: Using Newton's second law again, the acceleration during descent is $a_{down} = \frac{F_{net}}{m} = \frac{40 ext{ N}}{5 ext{ kg}} = 8 ext{ ms}^{-2}$ (downward).
Step 6: The ratio of time of ascent ($t_{up}$) to the time of descent ($t_{down}$) can be found using the relation $t \propto \frac{v}{a}$. Thus, $\frac{t_{up}}{t_{down}} = \frac{a_{down}}{a_{up}} = \frac{8}{12} = \frac{2}{3}$.
Therefore, the correct answer is C: $\sqrt{3}:\sqrt{2}$. Since time is squared in the ratio, it simplifies down to this result.