$T(n)=8T(\frac{n}{2})+qn$
Using master's theorem, we'll compare this will the standard formula:
$T(n)=aT(\frac{n}{b})+\theta(n^klog^pn)$, we find that
$a=8, b=2, k=1,p=0$, and $a>b^k$
The Order of the recurrence becomes
$\theta(n^{log_ba})\rightarrow\theta(n^{log_28})=\theta(n^3)$
Option (C).