Let $V$ and $\rho$ denote the volume and density of the block, respectively. From Archimedes' Principle, the buoyant force due to the water is
$\begin{align*}\left(3/4 \right)\cdot V \cdot \left(1000\; \mathrm{kg}\mathrm{/}\mathrm{m}^3\right) \cdot g\end{align*}$
and the buoyant force to the oil is
$\begin{align*}\left(1/4 \right)\cdot V \cdot \left(800\; \mathrm{ kg}\mathrm{/}\mathrm{m}^3\right) \cdot g\end{align*}$
So, the total buoyant force is
$\begin{align*}\left(3/4 \right)\cdot V \cdot \left(1000 \; \mathrm{ kg}\mathrm{/}\mathrm{m}^3\right) \cdot g + \left(1/4 \rig...$
This must equal the weight of the block, $g\cdot V\cdot \rho$ , so
$\begin{align*}g\cdot V\cdot \rho = V \cdot g \cdot 950 \; \mathrm{ kg}\mathrm{/}\mathrm{m}^3\end{align*}$
Thus,
$\begin{align*}\rho = \boxed{950 \; \mathrm{ kg}\mathrm{/}\mathrm{m}^3}\end{align*}$
Therefore, answer (C) is correct.

Solution to 2008 Problem 31