Answer
in the year $2021$
Work Step by Step
The model we arrived at was
$A=4000(1.006)^{4t}$
(amount t years after November 2010)
We now want to find $t$ for which
$\left[\begin{array}{ll}
5200 & =4000(1.006)^{4t}\\
5200/4000 & =(1.006)^{4t}\\
\log(52/40) & =4t\cdot\log 1.006\\
t & =\dfrac{\log(52/40)}{4\log 1.006}
\end{array}\right]$
$t\approx 10.964$
So, this will happen during the $11th$ year after 2010,
in the year $2021$.
