Your reaction needs #"203 g"# of carbon tetrachloride.
Start with the balanced chemical equation
#CH_(4(g)) + "CCl"_(4(l)) -> 2CH_2Cl_(2(g))#
Notice that you have a #"1:1"# mole ratio between methane and carbon tetrachloride, which means that, for every 1 mole of the former, you'll need exactly 1 mole of the latter.
You can determine how many moles of methane you have by using its molar mass
#"21.2 g CH"_4 * "1 mole"/"16.04 g" = "1.322 moles CH"_4#
This means that the number of moles of carbon tetrachloride needed is
#"1.322 moles CH"_4 * "1 mole CCl"_4/"1 mole CH"_4 = "1.322 moles CCl"_4#
Now use carbon tetrachloride's molar mass to calculate how many grams are needed
#"1.322 moles CCl"_4 * "153.82 g"/"1 mole" = "203.35 g"# #"CCl"_4#
Rounded to three sig figs, the answer will be
#m_("CCl"_4) = "203 g"#
