Let $\mathbf{a}=x \hat{i}+y \hat{j}+z \hat{k}, \mathbf{b}=3 \hat{i}+6 \hat{j}+6 \hat{k}$
$\mathbf{a} \cdot \mathbf{b}=27$
According to question, $\mathbf{a}$ is a collinear with $\mathbf{b}$, then
$\mathbf{a}=\lambda \mathbf{b}$ ...(i)
$\Rightarrow \quad \mathbf{a} \cdot \mathbf{b}=27$
$\Rightarrow \quad \lambda \mathbf{b} \cdot \mathbf{b}=27$ [from Eq. (i)]
$\Rightarrow \quad \lambda|\mathbf{b}|^2=27$
$\lambda=\frac{27}{\left(\sqrt{(3)^2+(6)^2+(6)^2}\right)^2}=\frac{27}{(9)^2}$
$\lambda=\frac{1}{3}$
$\mathbf{a}=\frac{1}{3}(3 \hat{i}+6 \hat{j}+6 \hat{k}) \Rightarrow \mathbf{a}=\hat{i}+2 \hat{j}+2 \hat{k}$
So, $|\mathbf{a}|=\sqrt{1+(2)^2+(2)^2}$
$=\sqrt{9}=3$ units