Let $f\left(x\right)=\left\{\begin{array}{l}|x-2|+a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x\le 2\\ 4x^2+3x+1,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ if }x>2\end{array}\right.$

If $f\left(x\right)$ has a local minimum at $x=2\text{, }$then

        Since $f$has a local minimum at $x = 2$

So   $f\left(2\right)\le f\left(x\right)$ for x in an open interval around 2.

$\Rightarrow f\left(2\right)\le \underset{x\to 2+}{lim} f\left(x\right)$

Thus,   $a=f\left(2\right)\le \underset{x\to 2}{lim} 4x^2+3x+1=21$