hint: 1 inch=25,4 MN, resistivity of aluminium=2,825×10^-6 cm ohms
82.7 cm or 32.5 in
The required resistance is calculated from the relationship of power, voltage, and resistance. The wire length to give that resistance is calculated from ...
L = A·R/ρ
The power is the ratio of energy to time:
P = E/t = (8 J)/(0.25 s) = 32 W
The required resistance is ...
R = V²/P = (240 V)²/(32 W) = 1800 Ω
Then the required wire length is ...
L = (π/4)(0.00016 in × 2.54 cm/in)²×(1800 Ω)/(2.825×10^-6 Ω·cm)
L ≈ 82.65 cm
The length of aluminum wire required is about 82.7 cm or 32.5 in.
This wire is roughly 66 AWG, about 1/6 the diameter of a human hair. Its melting current in still air is likely well below the current it carries in this heater. That is, it probably needs to experience significant airflow in order to maintain its integrity.