Answer
$$2.08 \times 10^{-6}$$
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
$$f(x)=\sqrt{x}\ \to\ f^{\prime \prime \prime}(x)=\frac{3}{8 x^{\frac{3}{2}}}$$
For $x=3.9 $, choose $K = \frac{3}{8 (3.9)^{\frac{3}{2}}}=0.0125$ and
\begin{aligned}
\left|f(3.9)-T_{2}(3.9)\right| & \leq(0.0125) \frac{|3.9-4|^{2+1}}{(2+1) !} \\
\left|f(3.9)-T_{2}(3.9)\right| & \leq(0.0125) \frac{0.001}{3 !} \\
&=(0.0125) \frac{0.001}{6}\\
& \approx 2.08 \times 10^{-6}
\end{aligned}
