Step-by-step Solution:

1. Recall the formula for the derivative of an inverse function. The formula for the derivative of the inverse of a function $f$ at a point $y_0$ is given by: $\left( {{f^{ - 1}}} \right)'\left( {{y_0}} \right) = \frac{1}{{f'\left( {{x_0}} \right)}}$ where $y_0 = f(x_0)$.

2. Identify the values for our problem. We need to find $\left( {{f^{ - 1}}} \right)'\left( 2 \right)$. So, $y_0 = 2$. We first need to find the value of $x_0$ such that $f(x_0) = 2$.

3. Find $x_0$ such that $f(x_0) = 2$. The given relation is: ${e^{ - x}}f\left( x \right) = 2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,\,dt}$ Let's substitute $f(x_0) = 2$ into this equation: ${e^{ - {x_0}}} \cdot 2 = 2 + \int\limits_0^{{x_0}} {\sqrt {{t^4} + 1} \,\,dt}$ Let's test a simple value for $x_0$, for example, $x_0 = 0$. If we set $x_0 = 0$, the integral term becomes $\int\limits_0^0 {\sqrt {{t^4} + 1} \,\,dt} = 0$. The equation becomes: ${e^0} \cdot 2 = 2 + 0$ $1 \cdot 2 = 2$ $2 = 2$ This is true. Therefore, $x_0 = 0$ is the value such that $f(0) = 2$.

4. Find the derivative of $f(x)$, i.e., $f'(x)$. We differentiate the given equation with respect to $x$: $\frac{d}{{dx}}\left( {{e^{ - x}}f\left( x \right)} \right) = \frac{d}{{dx}}\left( {2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,\,dt} } \right)$ For the left side, we use the product rule, $(uv)' = u'v + uv'$: $\frac{d}{{dx}}\left( {{e^{ - x}}f\left( x \right)} \right) = ( - {e^{ - x}})f(x) + {e^{ - x}}f'(x)$ For the right side, we use the Fundamental Theorem of Calculus, which states that $\frac{d}{{dx}}\int\limits_a^x {g(t)\,dt} = g(x)$: $\frac{d}{{dx}}\left( {2 + \int\limits_0^x {\sqrt {{t^4} + 1} \,\,dt} } \right) = 0 + \sqrt {{x^4} + 1}$ Equating the derivatives of both sides: $- {e^{ - x}}f(x) + {e^{ - x}}f'(x) = \sqrt {{x^4} + 1}$

5. Evaluate $f'(x_0)$ at $x_0=0$. We need to find $f'(0)$. Let's substitute $x=0$ into the differentiated equation: $- {e^{ - 0}}f(0) + {e^{ - 0}}f'(0) = \sqrt {{0^4} + 1}$ We know that $e^0 = 1$ and from Step 3, $f(0)=2$. $- (1)(2) + (1)f'(0) = \sqrt {0 + 1}$ $- 2 + f'(0) = \sqrt 1$ $- 2 + f'(0) = 1$ $f'(0) = 1 + 2 = 3$

6. Calculate $\left( {{f^{ - 1}}} \right)'\left( 2 \right)$. Using the formula from Step 1 with $y_0=2$ and $x_0=0$: $\left( {{f^{ - 1}}} \right)'\left( 2 \right) = \frac{1}{{f'\left( 0 \right)}}$ Substitute the value of $f'(0)$ we found: $\left( {{f^{ - 1}}} \right)'\left( 2 \right) = \frac{1}{3}$

7. Conclusion The value of $\left( {{f^{ - 1}}} \right)'\left( 2 \right)$ is $\frac{1}{3}$. This corresponds to option B.