Given $Adj\left(A\right)=k{A}^T.$

$\Rightarrow \left|Adj\left(A\right)\right|=\left|k{A}^T\right| ...\left(i\right)$

We know that, for a square matrix of order $n, \left|Adj\left(A\right)\right|={\left|A\right|}^{n-1}, \left|kA\right|=k^n\left|A\right|$ and $\left|A^T\right|=\left|A\right|.$

Hence, from equation $\left(i\right),$ we have   

$\Rightarrow {\left|A\right|}^{3-1}=k^3\left|A\right|$

$\Rightarrow {\left|A\right|}^2=k^3\left|A\right|$

$\Rightarrow k^3=\left|A\right|=27$

$\Rightarrow k=3.$

Hence, $k^2-3k+5=3^2-3\times 3+5=5.$