Answer
$0.78\;\rm J/K$
Work Step by Step
To estimate the net change in entropy of the system (water+aluminium), we need to find the mixture's final temperature.
We assume that the system is isolated since both of them are inside a Styrofoam container.
This means that the heat lost by the aluminum piece is the heat gained by the water.
Thus,
$$Q^-_{Al}=Q^+_{water}$$
$$m_{Al}c_{Al}\left(T_{i,Al}-T_{f}\right)=m_{water}c_{water}\left(T_{f}-T_{i,water}\right)$$
Solving for $T_f$;
$$m_{Al}c_{Al} T_{i,Al}-m_{Al}c_{Al} T_{f} =m_{water}c_{water}T_{f}-m_{water}c_{water}T_{i,water} $$
$$m_{Al}c_{Al} T_{i,Al} +m_{water}c_{water}T_{i,water} =m_{water}c_{water}T_{f}+m_{Al}c_{Al} T_{f}$$
$$T_f\left(m_{water}c_{water} +m_{Al}c_{Al}\right)=m_{Al}c_{Al} T_{i,Al} +m_{water}c_{water}T_{i,water}$$
$$T_f=\dfrac{m_{Al}c_{Al} T_{i,Al} +m_{water}c_{water}T_{i,water}}{m_{water}c_{water} +m_{Al}c_{Al}}$$
Plugging the known;
$$T_f=\dfrac{\left(2.8\cdot900\cdot 28.5\right) +\left(1\cdot 4186\cdot 20\right)}{(1\cdot 4186)+(2.8\cdot 900) }$$
$$T_f=\color{blue}{23.2}^\circ \rm C$$
Now we can calculate the entropy of the system but remember that the temperature of the mixture is not constant all the time, so we need to take the average temperature of each one of them.
Thus,
$$T_{avg,Al}=\dfrac{23.2+28.5}{2}=25.85^\circ \rm C\approx \color{blue}{299.01}\;K$$
$$T_{avg,water}=\dfrac{20+23.2}{2}=21.6^\circ \rm C\approx \color{blue}{294.76}\;K$$
The entropy change of the system:
$$\Delta S_{sys}=\Delta S_{Al}+\Delta S_{water}=\dfrac{Q^-_{Al}}{T_{avg,Al}}+\dfrac{Q^+_{water}}{T_{avg,water}}$$
$$\Delta S_{sys} =\dfrac{m_{Al}c_{Al}(T_{f,Al}-T_{i,Al})}{T_{avg,Al}}+\dfrac{m_{water}c_{water}(T_{f,water}-T_{i,water})}{T_{avg,water}}$$
Plugging the known;
$$\Delta S_{sys} =\dfrac{2.8\cdot 900\cdot (23.2-28.5)}{299.01}+\dfrac{1\cdot 4186\cdot(23.2-20)}{294.76}$$
$$\boxed{\Delta S_{sys}=\color{red}{\bf0.78}\;\rm J/K}$$
