$In\,{1^{st\,}}\,situation$

${V_b}{\rho _b}g = {V_s}{\rho _w}g$

$\frac{{{V_s}}}{{{V_b}}} = \frac{{{\rho _b}}}{{{\rho _w}}} = \frac{4}{5}$                  $…\left( i \right)$

Here $V_b$ is volume of block

$V_s$ is submerged volume of block

${\rho _b}$ is density of block

${\rho _w}$ id density of water $\& $ Let ${\rho _0}$ is density of oil 

Finally in equilibrium condition

${V_b}{\rho _b}g = \frac{{{V_b}}}{2}{\rho _0}g + \frac{{{V_b}}}{2}{\rho _w}g$

$2{\rho _b} = {\rho _0} + {\rho _w}$

$ \Rightarrow \frac{{{\rho _0}}}{{{\rho _w}}} = \frac{3}{5} = 0.6$