Yes. They are asking you to compute what is the length of $a$ after it has been projected on the axis that goes through the vector $b$.
Another way to state this : If you draw your vectors $\vec a$ and $\vec b$ on the plane, you can choose a vector that is orthogonal to $\vec b = (1.6,8.6)$, say for instance $\vec c = (-8.6,1.6)$ because in general a vector of the form $(x,y)$ is orthogonal to the vector $(-y,x)$ (it suffices to compute the scalar product to confirm that). After that, you want to add a certain multiple of this orthogonal vector to $\vec a$, to ensure that $\vec a + \lambda \vec c$ is parallel to $\vec b$, i.e, that you projected orthogonally $\vec a$ on $\vec b$. Say that this is done, i.e. that $ \vec a + \lambda \vec c = \mu \vec b $ then by computing the scalar product by $\vec b$ on both sides, you obtain $ \vec b \cdot \vec a = \vec b \cdot \vec a + \lambda (0) = \vec b \cdot \vec a + \lambda (\vec b \cdot \vec c)= \vec b \cdot (\vec a + \lambda \vec c) = \mu \vec b \cdot \vec b \quad \Longrightarrow \quad \mu = \frac{\vec b \cdot \vec a}{\vec b \cdot \vec b}. $ because by construction $\vec b \cdot \vec c = 0$. Now the component of $\vec a$ along the direction of $\vec b$ is precisely $ \mu \| \vec b \| = \frac{\vec b \cdot \vec a}{\vec b \cdot \vec b} \sqrt{ \vec b \cdot \vec b} = \frac{\vec b \cdot \vec a}{\sqrt{\vec b \cdot \vec b}}. $ Hope that helps!