Let each square be of side $x$. Then after removing squares of side $x$ from the sheet, we have a box of base $8-2x$, $3-2x$ and height $x$ left.

So volume of this box $ V = (8-2x)(3-2x)x = 8(3x-2x^2) - 2x(3x-2x^2) = 4x^3 - 22x^2 + 24x$

To find max volume, differentiate with respect to x and set to 0.

$ V' = 12x^2 - 44x + 24 = 0 $

$ 4(3x^2 - 11x + 6 ) = 0 $

$ (3x-2)(x-3) = 0 $

So $x = \frac{2}{3}$ or $x=3$

If $x=3$ then one side becomes $3-2x = -3$ which is not possible.

So $x=\frac{2}{3}$

Also differentiating V again ( to prove $x = \frac{2}{3}$ is point of maxima ),

$V'' = 4(6x-11) $

At $x=\frac{2}{3}$, $V'' = -28 < 0$

So $x=\frac{2}{3}$ is point of maxima.

So volume at $x=\frac{2}{3}$ is

$V = 4x^3 - 22x^2 + 24x = 4(\frac{2}{3})^3 - 22(\frac{2}{3})^2 + 24\frac{2}{3} = \frac{200}{27} $ cubic metre