Find the approximate change in the surface area

Let y be the surface area of the cube.

$y=6 x^{2}$

We have

$\frac{\Delta x}{x} \times 100=1$

Now,

$\frac{d y}{d x}=12 x$

$\Rightarrow \Delta y=d y=\frac{d y}{d x} d x=12 x \times \frac{x}{100}=0.12 x^{2} \mathrm{~m}^{2}$

Hence, approximate change in the surface area of the cube is $0.12 x^{2} \mathrm{~m}^{2}$.