A steel metre scale reads the correct length at 16^ ^ ^ C ....

The change in temperature is \Delta T = 46^{}^{{}^{\circ}} C - 16^{}^{{}^{\circ}} C = 30^{}^{{}^{\circ}} C.

When the temperature increases, the scale expands. The measured length $L'$ of an object of true length $L$ is given by:

$L' = L(1 - \alpha \Delta T)$

The error in measurement is:

$\Delta L = L' - L = -L \alpha \Delta T$

The percentage error is given by:

$\text{Percentage error} = \dfrac{\Delta L}{L} \times 100 = -\alpha \Delta T \times 100$

Substituting the given values \alpha = 11 \times 10^{-6} ^{}^{{}^{\circ}} C^{-1} and \Delta T = 30^{}^{{}^{\circ}} C:

$\text{Percentage error} = -11 \times 10^{-6} \times 30 \times 100 = -0.033\%$