c
\(In\,{1^{st\,}}\,situation\)

\({V_b}{\rho _b}g = {V_s}{\rho _w}g\)

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

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

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

\(2{\rho _b} = {\rho _0} + {\rho _w}\)

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