Answer
$$ 0.183$$
Work Step by Step
Since the error bound is given by
$$\left|T_{n}(x)-f(x)\right| \leq K \frac{|x-a|^{n+1}}{(n+1) !}$$
where $ \left|f^{(n+1)}(u)\right| \leq K$, then for
$$f(x)=e^x\ \to\ f^{(n)} (x)= e^x$$
For $x\in [0, 1.1] $ choose $K = e^{1.1}$ and
\begin{align*}
\left|e^{1.1}-T_{3}(1.1)\right| &\leq K \frac{(1.1-0)^{4}}{4 !}\\
&=\frac{e^{1.1} 1.1^{4}}{24}\\
& \approx 0.183
\end{align*}
