Determining asymptotic running time by profiling
If you remove the 200ms overhead, you get
$$
\begin{array}{c|ccccc}
n &amp; 2^{10} &amp; 2^{11} &amp; 2^{12} &amp; 2^{13} &amp; 2^{14} \\
t &amp; 100 &amp; 300 &amp; 900 &amp; 2700 &amp; 8100
\end{array}
$$
This shows that when $n = 2^{10+k}$, $t = 100 \cdot 3^k$. You take it from here. (You won't get $cn^{3/2}$, by the way! The setters made a mistake.)