Take into account that the change of the area can be modeled with the following expression:
[tex]\begin{gathered} A=4400(1-0.055)^t \\ A=4400(0.945)^t \end{gathered}[/tex]where A is the area and t the time.
In this case, replace t = 10 and simplify:
[tex]A=4400(0.945)^{10}\approx2499.02[/tex]Hence, after 10 years the area is approximately 2499.02 km^2