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.
