The answer is #"0.17 L"#, or #"170 mL"#.

You know you are dealing with a #"98%"# #"v/v"# solution of #H_2SO_4#. Since a #"v/v"# percent concentration expresses the volume of the solute divided by the total volume of the solution and multiplied by 100%, your initial #"98%"# #"v/v"# solutin will have, for every **100.0 mL**, **98.0 mL** of #H_2SO_4#. So,

#"98% v/v" = ("98.0 mL sulfuric acid")/("100 mL solution")#

SInce we know that the solution's density is #"1.80 g/mL"#, we can determine what mass of #H_2SO_4# we have for every *100 mL* of solution

#rho = m/V => m_(H_2SO_4) = rho * V = 1.80g/(mL) * 98.0mL#

#m_(H_2SO_4) = 176.4# #"g"#

The goal is to determine what the molarity of the concentrated sulfuric acid, and then use this value to perform a simple dilution calculation in order to calculate the volume needed.

Since molarity is expressed in moles of solute per volume of solution, you need the moles of #H_2SO_4#, which come from

#n_(H_2SO_4) = m/("molar mass") = ("176.4 g")/("98.0 g/mol") = 1.80# #"moles"#

This means that the molarity of the concentrated solution is

#C = n/V = ("1.80 moles")/("100" * 10^(-3) "L") = 18# #"M"#

The volume of concentrated solution needed is

#C_1V_1 = C_2V_2 => V_1 = C_2/C_1 * V_2#

#V_1 = ("0.5 M")/("18 M") * "6.00 L" = "0.17 L"#, or #V_1 = "170 mL"#