If $k$ is a scalar and $I$ is a unit matrix of order $3$, then $\text{adj}(kI)=$

We use the property $\text{adj}(kA)=k^{n-1}\text{adj}(A)$, where $n$ is the order of the square matrix $A$.
Here, $A=I$ and $n=3$.
$$\text{adj}(kI)=k^{3-1}\text{adj}(I)$$
Since the adjoint of an identity matrix is the identity matrix itself ($\text{adj}(I)=I$):
$$\text{adj}(kI)=k^{2}I$$