Answer
$n=5$
Work Step by Step
The error $R_n$ can be computed as:
$|R_n(0.6)|=|\dfrac{f^{n+1}(z)(x-c)^{n+1}}{(n+1)!}|$
Here, we have: $|R_n(0.6)|=\dfrac{e^{0.6}(0.6)^{n+1}}{(n+1)!}|$
or, $\dfrac{e^{0.6}(0.6)^{n+1}}{(n+1)!}\lt 0.001$
Now, we will use the trail and error method by plugging the different values of $n$.
Thus, $|R_2(2)|=\dfrac{e^{0.6}(0.6)^{2+1}}{(2+1)!} \approx 0.066$
$|R_3(3)|=\dfrac{e^{0.6}(0.6)^{3+1}}{(3+1)!} \approx 0.0098$
$|R_4(4)|=\dfrac{e^{0.6}(0.6)^{4+1}}{(4+1)!} \approx 0.0012$
$|R_5(5)|=\dfrac{e^{0.6}(0.6)^{5+1}}{(5+1)!} \approx 0.00012$
We can see that the error is less than $0.001$ for the value of $n=5$.
