Answer
$n=3$
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 $f(x)= \cos x$;
Then, we have $|f^{n+1}(c)|=\cos [(n+1) \dfrac{\pi}{2}+c]$
Thus, we have
$|R_n(x)| =(x-a)^{n+1}\dfrac{f^{n+1}(c)}{(n+1)!} =\dfrac{\cos [(n+1) \dfrac{\pi}{2}+c]}{(n+1)!} (x-a)^{n+1} $
or, $|R_n(x)|\leq |\dfrac{(0.25)^{n+1}}{(n+1)!}| \leq 10^{-3}$
or, $|\dfrac{(n+1)!}{(0.25)^{n+1}}| \leq 10^{3}$
or, $n=3$
