Ice and Water.

Let the edge at that moment be $a$ and height of water in that cylinder be $h$ .

Buoyancy Force on ice cube $=1×a^2h×g$

Weight of ice cube $=0.9×a^3×g$

Balancing the forces, we get $h=\dfrac{9a}{10}$

Now, we need one more equation, which is conservation of mass.

Initial mass of ice cube $=4^3×0.9$

Final mass of ice cube $=a^3×0.9$

Final mass of water $=(\pi r^2h-a^2h)×1$ , where 'r' is the radius of cylinder.

By conservation of mass,

$4^3×0.9=0.9a^3+9\pi ×0.9a-0.9a^3$

$a=\dfrac{576}{81\pi}=2.264$

First, I'd like to mention,- you don't have to go through this solution steps, if you wish so!

Just say- $side length,a={\frac{ {a_0}^3}{\pi r^2}}$

Its one of the many possible solution . But its the Easiest , undoubtedly! Because you don't need that density thing at all to solve it! How?

Assume,

Radius of the pot, $r={\frac{d}{2}}$

So Area of the pot, $\text{A}=\pi r^2$

Side length( Before ) = $a_0$

Exactly when the cube starts floating,

Side length = $a$

And Height of the submerged portion = $h$

Density of ice= $\rho_i$

Density of water= $\rho_w$

So,

Mass of previous cube = Mass of current cube + Mass of water

Water volume $=\text{A}h-a^2h$

( You might've understood how, right! )

So, $\rho_i{a_0}^3=\rho_ia^3+\rho_w h (\text{A}-a^2)$ $\rightarrow {a_0}^3-a^3={\frac{\rho_w}{\rho_i}} h (\text{A}-a^2)$

Then,

Mass of the current cube =Mass of $a^2 h$ Volume of water. $\rho_i a^3=\rho_w a^2 h$ $\rightarrow h={\frac{\rho_i}{\rho_w}} a$

Plug the value in previous equation- ${a_0}^3-a^3={\frac{\rho_w}{\rho_i}}h (\text{A}-a^2)$

$\rightarrow {a_0}^3-a^3={\frac{\rho_w}{\rho_i}}{\frac{\rho_i}{\rho_w}} a (\text{A}-a^2)$

$\rightarrow {a_0}^3-a^3=a(\text{A}-a^2)$

See, that density thing is gone!

So let's do the simple final job- ${a_0}^3-a^3=\text{A}a-a^3$ $\rightarrow a=\boxed{{\frac{ {a_0}^3}{A}}}$

Then the rest is putting the Values provided in the Problem . Its a piece of cake to you now! Its so simple, and excellently interesting , isn't it!

Hey it is a problem of HC Verma....