1. Let the initial velocity be $u$.

   Since the ball reaches maximum height $h$, $h=\frac{u^2}{2g}$ so, $u^2=2gh$

2. Use the equation of vertical motion for height $y=\frac{h}{3}$.

   The displacement after time $t$ is $y=ut-\frac{1}{2}gt^2$

   Put $y=\frac{h}{3}$: $ut-\frac{1}{2}gt^2=\frac{h}{3}$

   Using $h=\frac{u^2}{2g}$, $ut-\frac{1}{2}gt^2=\frac{1}{3}\cdot \frac{u^2}{2g}=\frac{u^2}{6g}$

3. Convert to a quadratic in $t$.

   Multiply by $6g$: $6gut-3g^2t^2=u^2$

   Rearranging, $3g^2t^2-6gut+u^2=0$

   Solve for $t$: $t=\frac{6gu\pm\sqrt{36g^2u^2-12g^2u^2}}{6g^2}$ $t=\frac{6gu\pm 2gu\sqrt{6}}{6g^2}$ $t=\frac{u}{g}\cdot \frac{3\pm \sqrt{6}}{3}$

   A simpler way is to use the standard result directly: $t=\frac{u\pm\sqrt{u^2-2gy}}{g}$

   At $y=\frac{h}{3}$, $u^2-2gy=u^2-2g\cdot \frac{h}{3}$ and since $u^2=2gh$, $u^2-2gy=2gh-\frac{2gh}{3}=\frac{4gh}{3}=u^2\left(1-\frac{1}{3}\right)=\frac{2u^2}{3}$ hence $\sqrt{u^2-2gy}=u\sqrt{\frac{2}{3}}$

   Therefore the two times are $t_1=\frac{u-u\sqrt{2/3}}{g}=\frac{u}{g}\left(1-\sqrt{\frac{2}{3}}\right)$ $t_2=\frac{u+u\sqrt{2/3}}{g}=\frac{u}{g}\left(1+\sqrt{\frac{2}{3}}\right)$

   Here:

   • $t_1$ = time while going up
   • $t_2$ = time while coming down

4. Find the required ratio.

   $\frac{t_1}{t_2}=\frac{1-\sqrt{2/3}}{1+\sqrt{2/3}}$

   Multiply numerator and denominator by $\sqrt{3}$: $\frac{t_1}{t_2}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

5. Match with options.

   This is Option B.

Final Answer: $\boxed{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}$