The heat dissipated in a resistance wire is given by Joule's law of...

The formula for heat dissipated is $H = I^2 R t$.

The maximum percentage error in $H$ is given by the sum of the percentage errors of its factors, multiplied by their respective powers:
$\dfrac{\Delta H}{H} \times 100 = 2 \left( \dfrac{\Delta I}{I} \times 100 \right) + \left( \dfrac{\Delta R}{R} \times 100 \right) + \left( \dfrac{\Delta t}{t} \times 100 \right)$

Given the percentage errors:
$\dfrac{\Delta H}{H} \times 100 = 9\%$
$\dfrac{\Delta R}{R} \times 100 = 2\%$
$\dfrac{\Delta t}{t} \times 100 = 1\%$

Substitute these values into the error equation:
$9\% = 2 \left( \dfrac{\Delta I}{I} \times 100 \right) + 2\% + 1\%$

$9\% = 2 \left( \dfrac{\Delta I}{I} \times 100 \right) + 3\%$

$2 \left( \dfrac{\Delta I}{I} \times 100 \right) = 9\% - 3\% = 6\%$

$\dfrac{\Delta I}{I} \times 100 = \dfrac{6\%}{2} = 3\%$

The percentage error in the measurement of current is $3\%$.

Answer: $3\%$