Center of gravity

Let $\overrightarrow{x}$ be the vector from the ground to the center of mass of the water.

Let $\overrightarrow{y}$ be the vector from the ground to the center of mass of the empty cube.

It is obvious that $\overrightarrow{y}$ has magnitude $1.5$ .

If $\overrightarrow{h}$ is the vector from the ground to the center of mass of the cube with the water in it, then $\overrightarrow{h}=\dfrac{a|\overrightarrow{y}|+b|\overrightarrow{x}|}{a+b}$ where a and b are the masses of the empty cube and water respectively.

The mass of the empty block is $a=\rho_{block} (3^3-2^3)=9500$

The mass of the water is $b=\rho_{water} (2^2)(|\overrightarrow{x}|-.5)(2)=8000(|\overrightarrow{x}|-.5)$

Plugging in for a and b

$|\overrightarrow{h}|=\dfrac{9500\times1.5+8000(|\overrightarrow{x}|-.5)|\overrightarrow{x}|}{9500+8000(|\overrightarrow{x}|-.5)}$

Using Wolfram, we get our maximum value as $M=|\overrightarrow{h}|=1.348\longrightarrow \lfloor 100M \rfloor=\boxed{134}$

$\text{Let k=1000. So mass of the cube = } k/2 * (3^3-2^3)=\dfrac {19k} 2.\\ Its ~distance ~ from ~ ground =\dfrac 3 2 ~m.\\ \text{Let water be filled up to x m from the ground}\\ Mass ~ of ~ water ~= k*(2*2*x - 2*2*\dfrac 1 2.)=4xk-2k.\\ Its ~ cg = \dfrac{x - \frac 1 2} 2 + \dfrac 1 2 = \dfrac{2x+1} 4 ~ m ~ from ~ ground.\\ \text{Taking moments about the ground }\\ c.g.= f(x)= \dfrac{\frac 3 2*\dfrac {19k}2 + \dfrac{2x+1} 4 *( 4xk-2k) } { \dfrac {19k}2 + ( 4xk-2k) }\\ = \dfrac{\dfrac {57k}4 + \dfrac{(2x+1)*(2x-1)k} 2} { ( \dfrac {19}2 +4x - 2)k } = \dfrac{55+8x^2}{30+16x}.\\ f ' (x)=\dfrac{16x(16x+30) - (8x^2+55)*16} {Can ~ be ~ any ~ thing !~ it ~is=(16x+30)^2} = 0. \\ \implies ~~ 8x^2+30x - 55=0 , ~solving ~for~ positive ~value ~ x= M=1.34 ~~\therefore ~100M=134.$