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Judging from some computer calculations $\lim_{n \rightarrow \infty} n/|{(a,b):2^a \cdot 3^b}| \rightarrow \infty$. This is the GAP code I used below; you can modify the initial value of n to quite a large value and it will still work.

n:=100000;;
count:=0;;
for a in [0..LogInt(n,2)] do
  for b in [0..LogInt(n,3)] do
    if(2^a*3^b<=n) then count:=count+1;;
  od;
od;
Print(Int(n/count),"\n");

Edit: using the above code I searched Sloane's website and uncovered this. Specifically, these are called 3-smooth numbers. There's formulae there too.

show/hide this revision's text 1

Judging from some computer calculations $\lim_{n \rightarrow \infty} n/|{(a,b):2^a \cdot 3^b}| \rightarrow \infty$. This is the GAP code I used below; you can modify the initial value of n to quite a large value and it will still work.

n:=100000;;
count:=0;;
for a in [0..LogInt(n,2)] do
  for b in [0..LogInt(n,3)] do
    if(2^a*3^b<=n) then count:=count+1;;
  od;
od;
Print(Int(n/count),"\n");