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

The true length of an object is $L$.

The scale is calibrated at T_1 = 16^{}^{{}^{\circ}} C. At T_2 = 6^{}^{{}^{\circ}} C, the change in temperature is \Delta T = 6^{}^{{}^{\circ}} C - 16^{}^{{}^{\circ}} C = -10^{}^{{}^{\circ}} C.

Due to thermal contraction, the length of one division on the scale becomes $d' = d(1 + \alpha \Delta T)$.

The measured length $L_m$ is given by $L_m = \dfrac{L}{1 + \alpha \Delta T} \approx L(1 - \alpha \Delta T)$.

The error in measurement is $\Delta L = L_m - L = -L \alpha \Delta T$.

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

Substituting the given values:

Percentage error $= -(11 \times 10^{-6}) \times (-10) \times 100 = 11 \times 10^{-3} = 0.011\%$.