Equate the volume of the hemisphere to the volume of the water displaced in the cylinder.

The volume of water displaced is equal to the volume of the hemisphere.
Volume of hemisphere ($V_h$) = $\frac{2}{3}\pi r_h^3$, where $r_h$ is the radius of the hemisphere.
Volume of water displaced in cylinder ($V_c$) = $\pi R_c^2 h_w$, where $R_c$ is the radius of the cylinder and $h_w$ is the height the water level rose.

$$V_h = V_c$$ $$\frac{2}{3}\pi r_h^3 = \pi R_c^2 h_w$$

Given $R_c = 10.5$ cm and $h_w = 6.2$ cm.

$$\frac{2}{3}\pi r_h^3 = \pi (10.5)^2 (6.2)$$

Cancel $\pi$ from both sides:

$$\frac{2}{3} r_h^3 = (10.5)^2 (6.2)$$ $$\frac{2}{3} r_h^3 = 110.25 \times 6.2$$ $$\frac{2}{3} r_h^3 = 683.55$$

Multiply both sides by $\frac{3}{2}$:

$$r_h^3 = \frac{683.55 \times 3}{2}$$ $$r_h^3 = \frac{2050.65}{2}$$ $$r_h^3 = 1025.325$$

Take the cube root of both sides:

$$r_h = \sqrt[3]{1025.325}$$ $$r_h \approx 10.0838 \text{ cm}$$

$$r_h \approx 10.1 \text{ cm}$$

The radius of the hemisphere is $\boxed{10.1 \text{ cm}}$.

16. Complete the figure below to show rotational order 3.

To complete the figure to show rotational order 3, the final shape must look identical after a rotation of $360^\circ / 3 = 120^\circ$ about its center. The small dot in the figure indicates the center of rotation.

The given figure shows one "point" or "arm" of a three-pointed star. To complete it, you need to draw two more identical points, each rotated by $120^\circ$ from the previous one around the central dot.

The completed figure will be a three-pointed star, where each point is identical to the one shown and is positioned at $120^\circ$ intervals around the center.