[tex]\bf P=1110.9e^{kt}\quad
\begin{cases}
P=1200 &\textit{in thousands}\\
t=2 &\textit{in 2002}
\end{cases}\implies 1200=1110.9e^{k2}
\\\\\\
\cfrac{1200}{1110.9}=e^{2k}\implies ln\left( \frac{1200}{1110.9} \right)=ln(e^{2k})\implies ln\left( \frac{1200}{1110.9} \right)=2k
\\\\\\
\cfrac{ln\left( \frac{1200}{1110.9} \right)}{2}=k\implies 0.0385755\approx k\implies 0.0386\approx k\\\\
-------------------------------\\\\[/tex]
[tex]\bf P=1110.9e^{0.0386t}\qquad
\begin{cases}
t=14\\
\textit{year 2015}
\end{cases}\implies P=1110.9e^{0.0386\cdot 14}
\\\\\\
P\approx 1907.0747\implies about\ 1,907,075\textit{ once rounded up}[/tex]
