Finding mass percent through molality of potassium nitrate solution Question

An aqueous $\ce{KNO3}$ solution has a molality of $\pu{4.16 m}$ and a density of $\pu{1.08 g/mL}$.  Calculate the percentage by mass $\ce{KNO3}$ of the solution.

My Approach
I know that molality ($b$) is defined as:
$$b = \frac{n_\mathrm{solute}}{m_\mathrm{solvent}}$$
In order to the mass of solution I multiplied $\pu{(1.08 g/mL)}\times\pu{1 mL}$, since, $m_\mathrm{solution} = m_\mathrm{solvent} + m_\mathrm{solute}$
However, I don't know where to use this information. I am also not sure what I to do with the $\pu{4.16 m}$ since I don't have moles of solute or mass of solvent. I am thinking that I have to assume some number here, but I might be wrong about this. Any help will be appreciated.
I am not sure you even need the density in this case.
Say you have $\ce{1kg}$ or $\ce{1000g}$ water (the solvent). That would mean that there is $\ce{4.16 mol}$ of $\ce{KNO_3}$.
Since we can figure out the molar mass of $\ce{KNO_3}$ = $\pu{101.11g mol-1}$, the mass of $\ce{KNO_3}$ in the sample must be $(4.16)(101.11)= \pu{420.62 g}$.
We already assumed that we had $\ce{1000g}$ of water, the mass percent of $\ce{KNO_3}$ is then equal to: $$\frac{420.62}{420.62+1000
} \approx 29.6\%$$