Let's mine for factors!

If a positive integer $n$ has 10 positive divisors, then it must be of the form $n = p\cdot q^4\ \ \ \ \text{or}\ \ \ \ n = p^9.$ where $p$ and $q$ are distinct prime numbers. The factors will be $1, q, q^2, q^3, q^4, p, pq, pq^2, pq^3, pq^4;\ \ \ \ \ \text{or}\ \ \ \ \ 1, p, p^2, \dots, p^9.$

The 12 smallest numbers of this form are $3\cdot 2^4 = 48 \\ 5\cdot 2^4 = 80 \\ 7\cdot 2^4 = 112 \\ \color{#3D99F6}{2\cdot 3^4 = 162} \\ 11\cdot 2^4 = 176 \\ 13\cdot 2^4 = 208 \\ 17\cdot 2^4 = 272 \\ 19\cdot 2^4 = 304 \\ 23\cdot 2^4 = 368 \\ \color{#3D99F6}{5\cdot 3^4 = 405} \\ 29\cdot 2^4 = 464 \\ 31\cdot 2^4 = 496$ The sum is $(3+5+7+11+13+17+19+23+29+31)\cdot 2^4 \\ + (2 + 5)\cdot 3^4 = \boxed{3095}.$

The interesting thing is (and I believe that is what Joel Yip was really after) that the next number with 10 positive divisors is $2^9 = 512$ .