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"
