1. Let the height of the tower be $h$ and the projection speed be $u$.
   Take downward direction as positive, so acceleration is $+g$.

2. Case 1: Projected vertically upward
   Since upward is opposite to our positive direction, initial velocity is $-u$.
   Displacement to the ground is $h$.

So, $h=-ut_1+\frac{1}{2}gt_1^2 \quad ...(1)$

3. Case 2: Projected vertically downward
   Initial velocity is now $+u$, and displacement is again $h$.

Thus, $h=ut_2+\frac{1}{2}gt_2^2 \quad ...(2)$

4. Case 3: Dropped from the top
   Initial velocity is $0$. Let the required time be $t$.

Then, $h=\frac{1}{2}gt^2 \quad ...(3)$

5. Use equations (1) and (2) to eliminate $u$.

From (1) and (2), since both equal $h$, $-ut_1+\frac{1}{2}gt_1^2=ut_2+\frac{1}{2}gt_2^2$

Rearrange: $u(t_1+t_2)=\frac{1}{2}g(t_1^2-t_2^2)=\frac{1}{2}g(t_1-t_2)(t_1+t_2)$

Hence, $u=\frac{g}{2}(t_1-t_2)$

6. Substitute into one of the equations for $h$. Using (2): $h=ut_2+\frac{1}{2}gt_2^2$ $h=\frac{g}{2}(t_1-t_2)t_2+\frac{1}{2}gt_2^2$ $h=\frac{g}{2}t_1t_2$

7. Now compare with the dropped case. From (3), $h=\frac{1}{2}gt^2$

So, $\frac{1}{2}gt^2=\frac{g}{2}t_1t_2$ $t^2=t_1t_2$ $t=\sqrt{t_1t_2}$

8. Match with the options.
   The correct option is: $\boxed{\sqrt{t_1t_2}}$ which is Option B.