As Mattos already said the mistake is that she adds $2000$ dollars at the end of each year.
After $1$ year she will have $2000$ dollars. After two years she will have $2000\cdot1.15+2000$, After three she will have $2000\cdot1.15^2+2000\cdot1.15+2000$.
After $n+1$ years she will have $2000(1.15^n+1.15^{n-1}\dots +1)$ By formula of geometric sum this is equal to: $2000(\frac{1.15^{n+1}-1}{1.15-1})$.
So after $n$ years she will have $2000(\frac{1.15^{n}-1}{0.15}$) So we must solve:
$1,000,000\cdot0.15=2000(1.15^{n}-1)\iff 500\cdot0.15=(1.15^{n}-1)\iff 76=1.15^n$ which has an approximate solution of $n=30.9865$
Therefore she will have $22+31=53$ years