Answer
$-56.4^{\circ} \rm C$
Work Step by Step
Let's assume that our system consists of aluminum, copper, and ethyl alcohol and this system is an isolated system.
And we know that the final temperature of the whole system is 25$^\circ$C.
Thus,
$$Q_{Al}+A_{Cu}+Q_{ethyl}=0$$
$$m_{Al}c_{Al}\Delta T_{Al}+m_{Cu}c_{Cu}\Delta T_{Cu}+m_{ethyl}c_{ethyl}\Delta T_{ethyl}=0$$
We have the mass of aluminum and copper but not the mass of ethyl.
So, $m_{ethyl}=\rho_{ethyl} V_{ethyl}$;
$$m_{Al}c_{Al} (T_f-T_{Al})+m_{Cu}c_{Cu}(T_f-T_{Cu})+\rho_{ethyl} V_{ethyl}c_{ethyl} (T_f-T_{ethyl})=0$$
Solving for $T_{Cu}$;
$$ m_{Cu}c_{Cu}(T_f-T_{Cu})=-m_{Al}c_{Al} (T_f-T_{Al})-\rho_{ethyl} V_{ethyl}c_{ethyl} (T_f-T_{ethyl}) $$
$$ T_f-T_{Cu} =\dfrac{ m_{Al}c_{Al} (T_{Al}-T_f )+\rho_{ethyl} V_{ethyl}c_{ethyl} ( T_{ethyl}-T_f) }{m_{Cu}c_{Cu}}$$
$$ T_{Cu} = T_f-\dfrac{ m_{Al}c_{Al} (T_{Al}-T_f )+\rho_{ethyl} V_{ethyl}c_{ethyl} ( T_{ethyl}-T_f) }{m_{Cu}c_{Cu}}$$
Plugging the known and do not forget to convert the temperatures to Kelvin.
$$ T_{Cu} = (25+273)-\dfrac{ (0.01)(900) (200-25)+(790)(50\times 10^{-6}) (2400)( 15-25) }{(0.02)(385)}$$
$$T_{Cu} =\color{red}{\bf 216.6}\;\rm K\approx \color{red}{\bf -56.4^\circ} \rm C$$