Which of these would neutralize the most #"HCl"#?

Likely #"Mg"("OH")_2#, as although both #A# and #B# are poorly soluble in PURE water (#"0.0001 g/100 mL"# vs. #"0.00064 g/100 mL"# at #25^@ "C"#), aluminum hydroxide is an acid.

Well, if you want to neutralize an acid, you need a base... And the best base supplies the most strong base, namely, the most hydroxide.

#"H"_2"CO"_3# is a weak acid (what is its name?), and thus, its conjugate base is a weak base.

#"H"_2"CO"_3(aq) rightleftharpoons "HCO"_3^(-)(aq)+ "H"^(+)(aq)#

And its conjugate base is a stronger weak base.

#"HCO"_3^(-)(aq) rightleftharpoons "CO"_3^(2-)(aq)+ "H"^(+)(aq)#

Neither #"HCO"_3^(-)# nor #"CO"_3^(2-)# would neutralize strong acid nearly as well as a strong base would... in fact, it would form a buffer that resists the #"HCl"#...

And so, the only reasonable options are between #A# or #B#.

In actuality, #"Al"("OH")_3# is a Lewis acid, NOT a base. In basic solution, it forms #"Al"("OH")_4^(-)#, because it accepts electrons from #"OH"^(-)# in solution. In #"HCl"# it becomes more soluble, just like #"Mg"("OH")_2# would.

Therefore, we say that #B# is correct, based on the solubility of #"Mg"("OH")_2# in water vs. that of #"Al"("OH")_3# in water, and the fact that aluminum hydroxide is an acid...