Answer
The density of the rock is $2.99\times 10^3~kg/m^3$
Work Step by Step
Let $M_a$ be the apparent weight of the rock. Let $\rho_w$ be the density of water. We can find the volume $V$ of the rock.
$M_a~g = M~g-F_B$
$F_B = M~g-M_a~g$
$\rho_w~V~g = M~g-M_a~g$
$V = \frac{M-M_a}{\rho_w}$
$V = \frac{9.28~kg-6.18~kg}{1.00\times 10^3~kg/m^3}$
$V = 3.10\times 10^{-3}~m^3$
We can find the density $\rho_r$ of the rock as:
$\rho_r = \frac{9.28~kg}{3.10\times 10^{-3}~m^3}$
$\rho_r = 2.99\times 10^3~kg/m^3$
The density of the rock is $2.99\times 10^3~kg/m^3$.
