Answer
$0.973\;\rm kg$
Work Step by Step
$\bullet\bullet$We have here 5 stages of heating for the ice:
1) Heating the ice from -10$^\circ$C to 0$^\circ$C, and the heat needed for that is given by
$$Q_1=m_{H_2O} c_{ice}\Delta T_{ice}$$
2) Melting the ice at 0$^\circ$C until all of it became water, and the heat needed for that is given by
$$Q_2=m_{H_2O}L_{{\rm f},ice}$$
3) Heating the water (the ice after melting became water) from 0$^\circ$C to 20$^\circ$C, and the heat needed for that is given by
$$Q_3=m_{H_2O}c_{water}\Delta T_{water}$$
$\bullet\bullet$ We have one stage of heating for the aluminum bucket which is cooling it from 70$^\circ$C to 20$^\circ$C. This amount of heat is given by
$$Q_4=m_{Al}c_{Al}\Delta T_{Al}$$
Let's assume that this system [ice+aluminum] is an isolated one.
Hence,
$$Q_1+Q_2+Q_3+Q_4=0$$
Plugging from above and then solving for $m_{Al}$;
$$m_{H_2O} c_{ice}\Delta T_{ice}+m_{H_2O}L_{{\rm f},ice}+\\m_{H_2O}c_{water}\Delta T_{water}+m_{Al}c_{Al}\Delta T_{Al}=0$$
$$m_{Al}=\dfrac{m_{H_2O} c_{ice}\Delta T_{ice}+m_{H_2O}L_{{\rm f},ice}+ m_{H_2O}c_{water}\Delta T_{water}}{-c_{Al}\Delta T_{Al}}$$
Plugging the known;
$$m_{Al}=\dfrac{(0.1) (2090)(10)+(0.1)(3.33\times 10^5)+ (0.1)(4190)(20)}{-(900)(-50)}$$
$$m_{Al}=\color{red}{\bf 0.973}\;\rm kg$$