According to the condition of floating, 

      $0.8{\mathrm{\rho }}_{\mathrm{w}}{\mathrm{gh}}_{\mathrm{k}}+{\mathrm{\rho }}_{\mathrm{w}}{\mathrm{gh}}_{\mathrm{w}}=0.9{\mathrm{\rho }}_{\mathrm{w}}\mathrm{gh}$      …(i)

where ${\mathrm{h}}_{\mathrm{k}}\text{ and }{\mathrm{h}}_{\mathrm{w}}$ are the submerged depth of the ice in the kerosene and water, respectively.

     ${\mathrm{h}}_{\mathrm{k}}+{\mathrm{h}}_{\mathrm{w}}=\mathrm{h}$               …(ii)

Solving Eqs. (i) and (ii), we get ${\mathrm{h}}_{\mathrm{k}}=0.5\mathrm{cm},{\mathrm{h}}_{\mathrm{w}}=0.5\mathrm{cm}$

$1{\mathrm{cm}}^3\text{ ice }\stackrel{\text{ melts }}{⟶}0.9{\mathrm{cm}}^3\text{ water }$

After melting the ice, $0.5{\mathrm{cm}}^3$ of volume will be occupied by the kerosene. Therefore, the fall in the level of kerosene is ${\mathrm{\Delta h}}_{\mathrm{K}}=\frac{0.5}{\mathrm{A}}$. Out of 0.9 cm³ of water formed, 0.5cm³ of water will be filled in the vacant part of ice and remaining $(0.9-0.5)=0.4{\mathrm{cm}}^3$ of water will be available to increase the level of water. 

Rise in the level of water is ${\mathrm{\Delta h}}_{\mathrm{w}}=\frac{0.9-0.5}{\mathrm{A}}=\frac{0.4}{\mathrm{A}}$

Hence, the net fall in the overall level is

       $\mathrm{\Delta h}=\frac{0.1}{\mathrm{A}}=\frac{0.1}{10}=0.01\mathrm{cm}=0.1\mathrm{mm}$