The percentage error in calculating the volume of a cubical box if an error of 1% is made in measuring the length of edges of the cube, is

3%

Let x be the length of an edge of the cube and y be its volume. Then, $y=x^3$.

Let $\Delta x$ be the error in x and $\Delta y$ be the corresponding error in y.

$\frac{\Delta x}{x} \times 100=1$               [Given]

$\Rightarrow \frac{d x}{x} \times 100=1$                       [∵  dx ≅  Δx]    ....(i)

We have to find $\frac{\Delta y}{y} \times 100$.

Now, $y=x^3 \Rightarrow \frac{d y}{d x}=3 x^2$

∴  $d y=\frac{d y}{d x} d x$

$\Rightarrow d y=3 x^2 . d x \Rightarrow \frac{d y}{y}=\frac{3 x^2}{y} d x \Rightarrow \frac{d y}{y}=\frac{3 x^2}{x^3} d x ~~~\left[∵ y=x^3\right]$

$\Rightarrow \frac{d y}{y}=3 \frac{d x}{x} \Rightarrow \frac{d y}{y} \times 100=3\left(\frac{d x}{x} \times 100\right)=3$        [Using (i)]

$\Rightarrow \frac{\Delta y}{y} \times 100=3$              $[∵ d y \cong \Delta y]$

So, there is 3% error in calculating the volume of the cube.