Convert the percentages to grams. If there were 100 g, each 1 % would equal 1 g.
So,
$"29.28 % C = 29.28g C"$
$"3.68 % H = 3.68 g H"$
$"39.01% O" = "39.01 g O"$
$"28.03% Na" = "28.03 g Na"$
Then convert the grams to moles
$ ("29.28 g")/("12.01 g/mol") = "2.438 mol C"$
$("3.68 g")/ ("1.008 g/mol") = "3.65 mol H"$
$("39.01 g")/("16.00 g/mol") = "2.438 mol O"$
$ ("28.03 g")/("22.99 g/mol")= "1.219 mol Na"$
Convert these numbers of moles to mole ratios by dividing all of them by the smallest mole number.
$ 2.438 /1.219 = 2.000/1 = "2 O: 1 Na"$
$ 2.438/ 1.219= 2.000/1 = "2 C : 1 Na"$
$ 3.65/1.219 = 2.99/1 = "3 H : 1 Na"$
So, the molar ratio is $"1 Na : 2 C : 2 O : 3 H"$,
and the empirical formula is
$"NaC"_2"H"_3"O"_2$