Let $f(x)=p(x) / q(x)$. Then, $x^{3} q(x)-p(x)$ has $1,2,3,4,5$ as roots. Therefore, WLOG, let $$x^{3} q(x)-p(x)=(x-1)(x-2)(x-3)(x-4)(x-5)=x^{5}-15 x^{4}+85 x^{3}-\ldots$$ Thus, $q(x)=x^{2}-15 x+85$, so $q(0)=85$. Plugging $x=0$ in the above equation also gives $-p(0)=-120$. Hence, the answer is $\frac{120}{85}=\frac{24}{17}$. Remark. From the solution above, it is not hard to see that the unique $f$ that satisfies the problem is $$f(x)=\frac{225 x^{2}-274 x+120}{x^{2}-15 x+85}$$