Let the determinant of a square matrix $A$ of order $m$ be $m-n$, where $m$ and $n$ satisfy $4m+n=22$ and…

We have been given that 

$\left|A\right|=m-n$, where $4m+n=22 .......\left(\mathrm{i}\right)$

and $17m+4n=93 .......\left(\mathrm{ii}\right)$

Solving $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$

$m=5, n=2$

Order $=5$

$\Rightarrow \left|A\right|=3$

We know that $\left|adjA\right|={\left|A\right|}^{w-1}$$\left|adj\left(adjA\right)\right|={\left|A\right|}^{{\left(w-1\right)}^2}$ where $w\times w$ is the order of the matrix.

$\therefore det \left(n adj \left(adj\left(mA\right)\right)\right)$

$=\left|2adj \left(adj \left(5A\right)\right)\right|$

$=2^5{\left|5A\right|}^{16}$

$=2^5 5^{80} {\left|A\right|}^{16}=2^5·3^{16}·5^{80}$

$=3^{11} 5^{80} 6^5$

So, $a+b+c=96$

Hence this is the required option.