Given two cubes of $1 \mathrm{m}$ side each and relative density $0.60$ and $1.15$ respectively.

This means that one block is lighter than water and the other one heavier than water. So the lighter one $(Block\, A)$ floats on water and the heavier one$(Block \,B)$ sinks fully. So in equilibrium the arrangement will be somewhat similar to the one shown in figure.

since the system is in equilibrium there is no net fore on the system.

Therefore entire upward force is equal to the total downward fore.

Total upward force $=$ Buoyancy force on $\mathrm{A}\left(B_{a}\right)+$

Buoyancy force on $\mathrm{B}\left(B_{b}\right)$ Buoyancy force $=$ density of liquid in which the object is immersed $\times$ volume of liquid displacedltimes gravitational constant.

$B_{a}=\rho \times((1-x) m \times 1 m \times 1 m) \times g$

$B_{b}=\rho \times(1 m \times 1 m \times 1 m) \times g$

Total buoyancy force $=B_{a}+B_{b}$

$=\rho \times g \times((2-x) m) \times(1 m \times 1 m)$       $….(1)$

Total downward force $=$ Weight of $\mathrm{A}\left(W_{a}\right)+$ Weight of $\mathrm{B}\left(W_{b}\right)$ Weight $=$ density of the block $\times$ volume of the block Itimes gravitational constant.

$W_{a}=\rho(A) \times(1 m \times 1 m \times 1 m) \times g$

$W_{b}=\rho(B) \times(1 m \times 1 m \times 1 m) \times g .$ Total downward force $=W_{a}+W_{b}$

$=\rho(A) \times(1 m \times 1 m \times 1 m) \times g+\rho(B) \times(1 m \times 1 m \times 1 m) \times g$

$=\rho(A)+\rho(B) \times(1 m \times 1 m \times 1 m) \times g$

$=(0.60+1.15) \times(1 m \times 1 m \times 1 m) \times g$

$=1.75 \times(1 m \times 1 m \times 1 m) \times g$         $…(2)$

Equations $1$ and $2$ are equal

$\Rightarrow \rho \times g \times((2-x) m) \times(1 m \times 1 m)=1.75 \times(1 m \times 1 m \times 1 m) \times g$

$\Rightarrow 2-x=1.75 \text { (since } \rho=1 \text { for water })$

$\Rightarrow \quad x=0.25 m(\text { or }) 25 \mathrm{cm}$

Therefore under equilibrium the lighter cube will project above the water surface to a height of $25 \mathrm{cm} .$