Answer
$n=4$
Work Step by Step
The remainder in the n-th order Taylor's polynomial $T_n(x)$ for a function $f(x)$ centered at $a$ can be expressed as:
$|R_n(x)|=|f(x)-T_n(x)| \leq (x-a)^{n+1}\dfrac{|x-a|^{n+1}}{(n+1)!}$
We are given that $f(x)= \ln (1+x)$;
Then, we have $|f^{n+1}(c)|=\dfrac{(-1)^n n!}{(1+x)^{n+1}}$
Thus, we have
$|R_n(x)| =(x-a)^{n+1}\dfrac{f^{n+1}(c)}{(n+1)!} =\dfrac{\dfrac{(-1)^n n!}{(1+x)^{n+1}}}{(n+1)!} (x-a)^{n+1} $
or, $\dfrac{(0.3)^{n+1}}{n+1} \leq 10^{3}$
or, $n=4$
