6. Given that $f(x)$ is an $n$-degree polynomial with non-negative integer coefficients, if $f(1)=6, f(7)=3438$, then $f(2)=$
6.43
Analysis: Let $f(x)=\sum_{i=0}^{n} a_{i} x^{i}$, where $a_{i} \in N(1 \leq i \leq n)$, since
$$
3438=f(7)=a_{0}+a_{1} 7+a_{2} 7^{2}+a_{3} 7^{3}+a_{4} 7^{4}+a_{5} 7^{5}+\mathrm{L}+a_{n} 7^{n}, 7^{5}>3438
$$

Thus, for $k \geq 5$, $a_{k}=0$, so $f(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}$. At this point, we have
$$
6=f(1)=a_{0}+a_{1}+a_{2}+a_{3}+a_{4} \quad, \quad 3438=f(7)=a_{0}+7 a_{1}+49 a_{2}+343 a_{3}+2401 a_{4} 。
$$

If $a_{4}=0$, then $a_{3} \leq 6, a_{0}=a_{1}=a_{2}=0$, at this time
$$
a_{0}+7 a_{1}+49 a_{2}+343 a_{3}+2401 a_{4}=343 \times 6=2058 \neq 3438$ ，

which contradicts $f(7)=3438$, so $a_{4}=1$. Therefore, we get
$$
\begin{array}{l}
a_{0}+a_{1}+a_{2}+a_{3}=6-a_{4}=6-1=5 \\
a_{0}+7 a_{1}+49 a_{2}+343 a_{3}=3438-2401 a_{4}=3438-2401=1037
\end{array}
$$

If $a_{3} \leq 2$, then $a_{3} \leq 2, a_{2} \leq 3, a_{0}=a_{1}=0$, at this time
$$
a_{0}+7 a_{1}+49 a_{2}+343 a_{3}=49 \times 3+343 \times 2=833 \neq 1037$ ，which would lead to a contradiction, so $a_{3}=3$. Hence
$$
a_{0}+a_{1}+a_{2}=5-a_{3}=5-3=2 \quad, \quad a_{0}+7 a_{1}+49 a_{2}=1037-343 a_{3}=1037-343 \times 3=8 \text { . }
$$

It is easy to see that to satisfy the above equations, only $a_{2}=0, a_{0}=a_{1}=1$.
So the polynomial is $f(x)=1+x+3 x^{3}+x^{4}$, at this time
$$
f(2)=1+2+3 \times 2^{3}+2^{4}=1+2+24+16=43 \text { . }
$$