Answer
$$0.0234375$$
Work Step by Step
Since
\begin{align*}
f(x)&=x^3\\
f'(x)&= 3x^2 \\
f''(x)&= 6x
\end{align*}
Then for $x\in [1,3] $, $|f''(x)|\leq 18$. Hence, $K=18$.
Thus:
\begin{align*}
\left|M_{16}-\int_{1}^{3} x^{3} d x\right| &\leq \frac{K \cdot 2^{3}}{24 \cdot 16^{2}}\\
&=\frac{18}{768}\\
&\approx 0.0234375
\end{align*}
