A block of wood floats in water with ${\left(\frac{4}{5}\right)}^{\mathrm{th}}$of its volume submerged. If the same block just floats in a liquid, the density of the liquid (in kg m^-3) is

According to Archimede’s principle Weight of the body = Weight of the liquid displaced 
Let V be volume of block
In water, ${\mathrm{V\rho }}_{\text{block }}\mathrm{g}=\left(\frac{4}{5}\mathrm{V}\right){\mathrm{\rho }}_{\text{water }}\mathrm{g}$
${\mathrm{\rho }}_{\text{block }}=\frac{4}{5}{\mathrm{\rho }}_{\text{water }}$      ….(i)
In liquid,${\mathrm{V\rho }}_{\text{block }}\mathrm{g}={\mathrm{V\rho }}_{\text{liquid }}\mathrm{g}$
${\mathrm{\rho }}_{\text{block }}={\mathrm{\rho }}_{\text{liquid }}$          …(ii)
From eqn. (i) and (ii), we get 
${\mathrm{\rho }}_{\text{block }}=\frac{4}{5}{\mathrm{\rho }}_{\text{water }}=\frac{4}{5}\mathrm{X}1000=800{\mathrm{kgm}}^{-3}$