Answer
$n=1$
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 $|f^{n+1}(c)|=\dfrac{(-1)^n (2n)!}{2^{2n+1}(1+x)^{2n+1/2}}$
Thus, we have
$|R_n(x)| =(x-a)^{n+1}\dfrac{f^{n+1}(c)}{(n+1)!} =\dfrac{\dfrac{(-1)^n (2n)!}{2^{2n+1}(1+x)^{2n+1/2}}}{(n+1)!} (x-a)^{n+1} $
or, $\dfrac{(2n)!(0.06)^{n+1} \times 10^3}{2^{2n+1} (n!)^2(n+1)} \leq 1$
or, $n=1$
