It is evident that $x^3 \leq 3x$ when $x \geq \sqrt{3}<2$. So it is better to maximize $a$ within the given conditions. We get $a=2^4\times3^2\times5\times 7$ while $b=11\times 13 = 143$ Required answer is $(3a+b) \pmod{1000}$. Notice that we only need to find $2^3 \times 3^3 \times 7 \pmod{100}$ by the "division property". Do the math and find that this is $\equiv 12 \pmod{100}$. So answer is $12\times 10 + 143=\boxed{263}$.