Answer
$\lt 1.87 \times 10^{-4}$
Work Step by Step
The Taylor series for $ e^x $ can be defined as: $ e^x=1+x +\dfrac{x^2}{2}+\dfrac{ x^3}{6}-....$
Consider the Remainder Estimation Theorem to find the error.
So, $|R_n(x)| \leq M \dfrac{|x-a|^{n+1}}{(n+1)!}$
and $|R_2(x)| \leq |\dfrac{e^ex^3}{3!}| $
So, Error $ \lt \dfrac{3^{0.1} \cdot (01)^{3} }{3!} \lt 1.87 \times 10^{-4}$
