A hollow spherical body with inner and outer radii of 6 cm and 8 cm...

Let the density of the material of the sphere be $\rho$.

Inner radius of the hollow sphere, $r = 6 \text{ cm}$.

Outer radius of the hollow sphere, $R = 8 \text{ cm}$.

Volume of the material of the hollow sphere is given by $V = \dfrac{4}{3}\pi(R^3 - r^3)$.

Weight of the hollow sphere is $W = V\rho g = \dfrac{4}{3}\pi(R^3 - r^3)\rho g$.

Since the sphere floats half-submerged in water, the volume of water displaced is half of the total volume of the sphere.

Volume of displaced water, $V_d = \dfrac{1}{2} \times \dfrac{4}{3}\pi R^3$.

Buoyant force exerted by water is $F_B = V_d \rho_w g = \dfrac{1}{2} \times \dfrac{4}{3}\pi R^3 \rho_w g$, where $\rho_w = 1000 \text{ kg/m}^3$ is the density of water.

By the principle of flotation, the weight of the sphere equals the buoyant force:

$\dfrac{4}{3}\pi(R^3 - r^3)\rho g = \dfrac{1}{2} \times \dfrac{4}{3}\pi R^3 \rho_w g$

$\rho = \dfrac{R^3}{2(R^3 - r^3)} \rho_w$

Substituting the given values:

$\rho = \dfrac{8^3}{2(8^3 - 6^3)} \times 1000$

$\rho = \dfrac{512}{2(512 - 216)} \times 1000$

$\rho = \dfrac{512}{2 \times 296} \times 1000$

$\rho = \dfrac{512}{592} \times 1000 \approx 864.86 \text{ kg/m}^3$

Rounding to the nearest integer, we get $865 \text{ kg/m}^3$.