It will be continuous at $x=2$ if $\lim\limits_{x\to 2}f(x)=f(2)$
$\Rightarrow \lim\limits_{x\to 2}\large\frac{x^3+x^2-16x+20}{(x-2)^2}$$=K$
$\Rightarrow K=\lim\limits_{x\to 2}\large\frac{(x-2)^2(x+5)}{(x-2)^2}$
$\quad\quad=\lim\limits_{x\to 2}(x+5)=7$
$K=7$
Hence (b) is the correct answer.