Step 1: Identify the temperature change.
The tape is calibrated at $T_1 = 25^\circ C$ but used at $T_2 = -15^\circ C$. The change is \[ \Delta T = T_2 - T_1 = -15 - 25 = -40^\circ C \]
Step 2: Recall the linear expansion relation.
The fractional change in length of a metal scale is \[ \frac{\Delta L}{L} = \alpha \Delta T \] where $\alpha = 1\times10^{-5}\ ^\circ C^{-1}$.
Step 3: Substitute the values.
\[ \frac{\Delta L}{L} = (1\times10^{-5})(-40) = -4\times10^{-4} \]
Step 4: Interpret the sign.
The negative sign means the tape contracts in the cold, so each marked division is slightly shorter than its labelled value, which introduces a measurement error.
Step 5: Convert the fractional change to a percentage.
\[ \text{percentage error} = \left|\frac{\Delta L}{L}\right|\times 100 = 4\times10^{-4}\times 100 \]
Step 6: Compute the final value.
\[ = 4\times10^{-2} = 0.04\% \] So the percentage error in the length measurement is $0.04\%$, matching option (1). \[ \boxed{0.04\%} \]