If A is a square matrix of order 3 such that adj (2A) = k adj (A),

For any matrtix $A$ of order $n$, adj $(\lambda A)=\lambda^{n-1}(\operatorname{adj} A)$, where $\lambda$ is a constant. Thus, for matrix $A$ of order 3, we have

$\operatorname{adj}(2 A)=2^{3-1}(\operatorname{adj} A)$

$\Rightarrow \operatorname{adj}(2 A)=2^{2}(\operatorname{adj} A)$

$\Rightarrow \operatorname{adj}(2 A)=4 \operatorname{adj}(A)$

$\Rightarrow k$ adj $(A)=4$ adj $(A) \quad[\because \operatorname{adj}(2 A)=k$ adj $(A)]$

$\Rightarrow k=4$