The points of maxima and minima can be found by differentiating the function w.r.t $x$ and equating it to zero.
$\therefore$ $f'(x)=6x^{2}-30x+36$
$\Rightarrow$ $6x^{2}-30x+36=0$ $\Rightarrow$ $x^{2}-5x+6=0$ $\Rightarrow$ $(x-2)\times(x-3)=0$
$\Rightarrow$ $x=2$ and $x=3$
Now, we have to find the value of $f''(x)$ at above points.
$\therefore$ $f''(x)=12x-30$
$\therefore$ $f''(2)=12(2)-30=-6$ $\Rightarrow$ at $x=2$ , function has maxima as $f''(2)<0$
Also, $f''(3)=12(3)-30=6$ $\Rightarrow$ at $x=3$, function has minima as $f''(3)>0$
Option C is correct.