The Peak

Did that orally... :-).
$x^2-5x+20\leq 0$ $\implies (x-4)(x-5)\leq 0$ $\implies x\in [4,5]$ $\implies M=\{x|x\in[4,5]\}$

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$f'(x)=6(x^2-5x+6)=(x-2)(x-3)$ We can draw sign scheme to see that $f'(x)>0$ in $M$ i.e $f(x)$ is increasing in $[4,5]$ . Hence maximum value of $f(x)$ occurs at $x=5$ .

$f(5)=2(5)^3-15(5)^2+36(5)-48$ $\huge \boxed{=7}$