Given,
The density of oil $\left({\mathrm{\rho }}_1\right)=724 \mathrm{kg}/{\mathrm{m}}^3$
The density of water $\left({\mathrm{\rho }}_2\right)=1000 \mathrm{kg}/{\mathrm{m}}^3$

According to Archimedes' principle,

Upthrust = Weight of the liquid displaced

$\Rightarrow {\mathrm{V\rho }}_{\mathrm{b}}\mathrm{g}={\mathrm{V}}_{\mathrm{o}}{\mathrm{\rho }}_1\mathrm{g}+{\mathrm{V}}_{\mathrm{w}}{\mathrm{\rho }}_2\mathrm{g}$   (${\mathrm{V}}_{\mathrm{o}}$ and ${\mathrm{V}}_{\mathrm{w}}$ are volumes of oil and water displaced respectively)

${\mathrm{V\rho }}_{\mathrm{b}}\mathrm{g}=\frac{\mathrm{V}}{6}\left(724\right)\mathrm{g}+\frac{5\mathrm{V}}{6}\left(1000\right)\mathrm{g}$, where ${\mathrm{\rho }}_{\mathrm{b}}$=density of the block.

and, ${\mathrm{V}}_{\mathrm{o}}=\frac{\mathrm{V}}{6}$ and ${\mathrm{V}}_{\mathrm{w}}=\frac{5\mathrm{V}}{6}$

$\Rightarrow {\mathrm{V\rho }}_{\mathrm{b}}\mathrm{g}=\left(\frac{724\mathrm{V}}{6}+\frac{5000\mathrm{V}}{6}\right)\mathrm{g}$
$\Rightarrow {\mathrm{V\rho }}_{\mathrm{b}}\mathrm{g}=\left(\frac{724\mathrm{V}+5000\mathrm{V}}{6}\right)\mathrm{g}$
$\Rightarrow {\mathrm{V\rho }}_{\mathrm{b}}\mathrm{g}=\frac{5724}{6}\mathrm{Vg}$

${\rho }_b=954\mathrm{ kg}/{\mathrm{m}}^3$