How to calculate the amount of aluminium that has been oxidised?
Initially, 1 kg of $\ce{Al}$ is present.  
After some time, oxidation increases the weight by 1%, or 10 g (e.g. 10 g of oxygen has been added).  
The atomic mass of oxygen is 16.0 g/mol.  
The molar ratio of $\ce{Al}$ to $\ce{O}$, from the molecular formula $\ce{Al2O3}$, is $\mathrm{\frac{2~mol~\ce{Al}}{3~mol~\ce{O}}}$.  


$$\mathrm{\frac{10~g~\ce{O}}{16.0~g~O/mol~O}=0.625~mol~O}$$  

$$\mathrm{\frac{2~mol~\ce{Al}}{3~mol~O}*0.625~mol~O=0.417~mol~\ce{Al}}$$  

$$\mathrm{\frac{27.0~g~\ce{Al}}{mol~\ce{Al}}*0.417~mol~\ce{Al}=11~g~\ce{Al}}$$

Note that even though only a single significant figure is given for each of steps 1 and 2 of the problem, I assumed that they should read "1.0 kg" and "1.0%" rather than just "1 kg" and "1%" for the purposes of illustrating the above calculation. Thereby, 0.42 mol or 11 g of aluminum was oxidized to $\ce{Al2O3}$.