Answer
$n=3$
Work Step by Step
$R_{n}(x) = \frac{f^{n+1}(z)}{(n+1)!} (x-c)^{n+1}$, where $z$ between $c$ and $x$.
$sin x$ will keep alternating between $sin x$, $cosx$, $-sinx$, and $-cosx$ no matter how many times you derive it.
$f^{n+1}(z) \leq 1$, for any $z$.
$c=0$ and $x=0.3$
$R_{n} (x) \lt 0.001$
$R_{n}(x) \leq \frac{1}{(n+1)!}(0.3)^{n+1}$
$\frac{1}{(n+1)!}(0.3)^{n+1} \lt 0.001$
By trail and error, $n=3$
