The question basically implies that $v_p(2018!)=3$. Now we use Legendre's formula. $v_p(2018!)$ =$\lfloor \frac{2018}{p} \rfloor$ +$\lfloor \frac{2018}{p^2} \rfloor$ +$\lfloor \frac{2018}{p^3} \rfloor$ +...and so on till $\infty$. Now if this sum exceeds $4$, that prime does not satisfy the $v_p$ condition. Now since our prime will be quite large, we can say that their squares and cubes and higher powers will definitely exceed $2018$. So all the terms from the second one onwards become =$0$. So we want that $\lfloor \frac{2018}{p} \rfloor$, must be equal to exactly 3. So $3p \leq 2018 <4p$. The smallest such prime is $p=\boxed{509}$. $\blacksquare$