Given, length of cubical block, $l=10 \mathrm{~cm}=0.1 \mathrm{~m}$






density of oil, $\rho_0=800 \mathrm{~kg} / \mathrm{m}^3$
and density of water, $\rho_w=10^3 \mathrm{~kg} / \mathrm{m}^3$
If $V_1$ and $V_2$ be the volume of wooden block in water and oil respectively, then
$$
\begin{aligned}
V_1: V_2 & =1.5: 8.5 \\
\frac{V_1}{V_2} & =\frac{1.5}{8.5}=\frac{3}{17} \\
\therefore \quad V_1 & =\frac{3}{17} V_2
\end{aligned}
$$
$\therefore V_1+V_2=V$, where, $V$ is the volume of the block.
From Eq. (i), we get
$$
\begin{array}{rlrl}
& \frac{3}{17} V_2+V_2 & =V \Rightarrow V_2=\frac{17}{20} \mathrm{~V} \\
\therefore \quad V_1 & =\frac{3}{20} \mathrm{~V}
\end{array}
$$

If $\rho$ be the density of the block.
$\therefore$ weight of block $=$ weight of oil displaced + weight of water displaced

$$
\begin{gathered}
V \rho G=V_1 \rho_\omega g+V_2 \rho_{\text {oil }} g \\
V \rho g=\frac{3}{20} V \rho_w g+\frac{17}{20} V \rho_{\text {oil }} g \\
\rho=\frac{3}{20} \rho_w+\frac{17}{20} \rho_{\text {oil }}=\frac{3}{20} \times 1000+\frac{17}{20} \times 800 \\
=830 \mathrm{~kg} / \mathrm{m}^3
\end{gathered}
$$
$\therefore$ Pressure difference
$$
p=\rho g h=830 \times 10 \times 0.1=830 \mathrm{~Pa}
$$