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 $|f^{n+1}(c)|=|e^{-c}| \leq 1$ for $c \in (0, 0.5)$ and $x=0.5; a=0$
Thus, we have
$|R_n(x)| =|\dfrac{(-1)^{n+1}e^{-c}(0.5)^{n+1}}{(n+1)!}| \leq |\dfrac{(0.5)^{n+1}}{(n+1)!}| $
or, $|\dfrac{(0.5)^{n+1}}{(n+1)!}| \leq 10^{-3}$
or, $n=4$
