How many mL of 6.0 M HCl solution will provide 164 g of HCl?

The idea here is that you need to determine how many moles you get in #"164 g"# of hydrochloric acid, then use the given solution's molarity to figure out the volume that would contain that many moles.

So, hydrochloric acid has a molar mass of #"36.46 g/mol"#, which means that the #"164-g"# sample will contain

#164color(red)(cancel(color(black)("g"))) * "1 mole HCl"/(36.46color(red)(cancel(color(black)("g")))) = "4.498 moles HCl"#

As you know, molarity is defined as moles of solute, which in this case is hydrochloric acid, divided by liters of solution

#color(blue)("molarity" = "moles of solute"/"liters of solution")#

A #"6.0-M"# hydrochloric acid solution will contain #6# moles of hydrochloric acid for every liter of solution. This means that #4.498# moles will come with a volume of

#c = n/V implies V = n/c#

#V = (4.498color(red)(cancel(color(black)("moles"))))/(6.0color(red)(cancel(color(black)("moles")))/"L") = "0.7497 L"#

Rounded to two sig figs, the number of sig figs you have for the molarity of the solution, the answer will be

#V = color(green)("0.75 L")#