Given,

$\left|A\right|=2$

And $n=\det \left(\underset{2024 \mathrm{times}}{\underbrace{adj\left(adj\left(adj...\left(A\right)\right)\right)}}\right)$

Now, using the formula $\underset{r \mathrm{times}}{\underbrace{\left|adj\left(adj\left(adj...\left(A\right)\right)\right)\right|}}={\left|A\right|}^{{\left(n-1\right)}^r}$ we get,

$\underset{2024 \mathrm{times}}{\left|\underbrace{adj\left(adj\left(adj...\left(A\right)\right)\right)}\right|}={\left|A\right|}^{{\left(n-1\right)}^{2024}}$

$\Rightarrow n={\left|A\right|}^{2^{2024}}$

$\Rightarrow n=2^{2^{2024}}$ $\left\{\text{as} \left|A\right|=2\right\}$

Now, solving

$2^{2024}=4^{1012}={\left(3+1\right)}^{1012}=3k+1$

$\Rightarrow 2^{2^{2024}}=2^{3k+1}$

$\Rightarrow 2^{2^{2024}}=2·2^{3k}$

$\Rightarrow 2^{2^{2024}}=2·{\left(9-1\right)}^k$

$\Rightarrow 2^{2^{2024}}=9m-2$

$\Rightarrow 2^{2^{2024}}=9t-2+9$

$\Rightarrow 2^{2^{2024}}=9m+7$

Hence, the remainder is $7$