What volume flask should be used to get 0.37 moles #I# present for every mole #I_2# at equilibrium?

Are we missing information? We need to be able to figure out the exact number of mols of #"I"# or #"I"_2#... no ratios, an actual number.

We begin with the ICE table:

#"I"_2(g) " "rightleftharpoons" " 2"I"(g)#

#"I"" "["I"_2]_i" "" "" "" "0#
#"C"" "-x" "" "" "+2x#
#"E"" "["I"_2]_i-x" "" "2x#

So we write the mass action expression to get:

#K_c = 0.011 = (["I"]^2)/(["I"_2])#

#= (2x)^2/(["I"_2]_i - x)#

What we apparently want is

#(2x)/(["I"_2]_i - x) = 0.37 = (["I"])/(["I"_2]) = ("mols I"//cancel"L")/("mols I"_2//cancel"L")#

We now have a system of equations:

#(2x)^2/(["I"_2]_i - x) = 0.011# #" "" "bb((1))#
(Note that #0.011# is in implied units of #"M"#.)

#(2x)/(["I"_2]_i - x) = 0.37# #" "" "bb((2))#
(Note that #0.37# is unitless.)

By inspection, we have:

#(2x)^2/(["I"_2]_i - x) = 0.011 = 0.37(2x) = 0.74x#

So,

#x = ("0.011 M")/0.74 = ul"0.0149 M"#

As a result, take #(2)# to get

#(2 cdot "0.0149 M")/(["I"_2]_i - "0.0149 M") = 0.37#

#2 cdot "0.0149 M" = 0.37(["I"_2]_i - "0.0149 M")#

#"0.0297 M" = 0.37["I"_2]_i - "0.0055 M"#

Therefore, the initial concentration of #"I"_2# is:

#color(blue)(["I"_2]_i) = ("0.0352 M")/(0.37) = color(blue)("0.0952 M")#

and the fraction of dissociation is

#alpha = x/(["I"_2]_i) = 0.1565#

Now we just need to know the ACTUAL mols of either #"I"# or #"I"_2#... the ratio desired can be ANY NUMBER of combinations of quantities...

The fraction of dissociation #alpha# and the concentration lost #x# also vary with concentration #["I"_2]_i#, so we cannot use those to determine the volume.