If this conclusion does not hold, let's assume $z$ is a root of $f(x)$ and $R(z) \ge 4$. If $R(z)\ge 4$, then $|z| \ge 4$ and $R\big(\frac{1}{z}\big) > 0$. it follows that: $$\Big{|}\frac{f(z)}{z^n}\Big| = \Big|a_0+\frac{a_1}{z} + \frac{a_2}{z^2} + \cdots + \frac{a_n}{z^n} \Big|$$ Because all $a_i$ are non-negative but not exceeding $9$, therefore we can derive the following relation by the triangular inequality: \begin{align} \Big{|}\frac{f(z)}{z^n}\Big| & \ge \Big|a_0 + \frac{a_1}{z}\Big| - \frac{9}{|z|^2} - \cdots -\frac{9}{|z|^n} \\ & > R\Big(a_0 + \frac{a_1}{z}\Big) - \frac{9}{|z|^2} - \cdots -\frac{9}{|z|^n} \\ & = R\Big(a_0 + \frac{a_1}{z}\Big) - 9\cdot\Big( \frac{1}{|z|^2} + \cdots + \frac{1}{|z|^n}\Big)\\ & = R\Big(a_0 + \frac{a_1}{z}\Big) - 9\cdot \frac{1}{|z|^2} \cdot \frac{1-\frac{1}{|z|^{n-1}}}{1-\frac{1}{|z|^2}}\\ &\ge 1- \frac{9}{|z|^2 -|z|} >0 \end{align} Therefore $z$ cannot be a zero. Contradiction.