The solution's molality is "5.0 molal".
So, you know that you are dealing with a "20%" by weight solution of "NaOH". In order to determine how much "NaOH" you actually have, you must use the solution's density, which is not given to you.
You could assume that you have a solution with a density that's a little bigger than water's, maybe "1.1 g/mL" or "1.2 g/mL", or you could look up the density of a 20% by weight sodium hidroxide solution, which is listed as "1.2191 g/mL".
Now, you know what the mass of the solution will be
rho = m/V => m_("solution") = rho * V = "1.2191 g/mL" * "550 mL"
m_("solution") = "670.5 g"
You can now find out how much "NaOH" you have
"w/w%" = m_("NaOH")/(m_("solution")) * 100 => m_("NaOH") = ("w/w" * m_("solution"))/100
m_("NaOH") = (20 * 670.5)/100 = "134 g"
Since molality is expressed in moles of solute divided by the mass of the solution - in kilograms, you'll need the number of "NaOH" moles
"134 g" * ("1 mole NaOH")/("40.0 g") = "3.35 moles"
Thus, the solution's molality is
"b" = ("3.35 moles")/("670.5" * 10^(-3)"kg") = "4.99 molal", or "5.0 molal" - rounded to two sig figs.
