The correct answer is $\boxed{\text{B}}$.
To find the value of $\left. {\frac{{{\text{dy}}}}{{{\text{dx}}}}} \right|_{{\text{x}} = 1}}$, we can use the following formula:
$$\frac{dy}{dx} = \lim_{h \to 0} \frac{y(x + h) – y(x)}{h}$$
In this case, we have $y(x) = x^2 + 2x + 10$, so we can substitute that into the formula to get:
$$\begin{align}
\frac{dy}{dx} &= \lim_{h \to 0} \frac{(x + h)^2 + 2(x + h) + 10 – (x^2 + 2x + 10)}{h} \
&= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 2x + 2h + 10 – x^2 – 2x – 10}{h} \
&= \lim_{h \to 0} \frac{2xh + h^2 + 2h}{h} \
&= \lim_{h \to 0} 2x + h + 2 \
&= 2x + 2
\end{align}$$
Evaluating at $x = 1$, we get:
$$\begin{align}
\frac{dy}{dx} &= 2(1) + 2 \
&= 4
\end{align}$$
Therefore, the value of $\left. {\frac{{{\text{dy}}}}{{{\text{dx}}}}} \right|_{{\text{x}} = 1} = \boxed{4}$.
The other options are incorrect because they do not represent the correct value of $\left. {\frac{{{\text{dy}}}}{{{\text{dx}}}}} \right|_{{\text{x}} = 1}$.