$\lim\limits_{x\to 2}f(x)=\lim\limits_{x\to 2}\large\frac{x^5-32}{x-2}$
$\Rightarrow \lim\limits_{x\to 2}\large\frac{x^5-2^5}{x-2}$
$\Rightarrow 5(2)^4$
$\Rightarrow 80$
Since $f(x)$ is continuous at $x=2$
$\lim\limits_{x\to 2}f(x)=f(2)$
$k=80$
Hence (b) is the correct answer.