The position of a particle as a function of time is given by $x=\left(5 t^{2}-4 t+5\right) \mathrm{m}$.

To find the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$, we first need to find the velocity of the particle as a function of time.

The velocity $v$ is the time derivative of the position $x$:

$$ v = \frac{dx}{dt} $$

Taking the derivative of $x$ with respect to $t$, we get:

$$ v = \frac{dx}{dt} = 10t - 4\,\mathrm{m/s} $$

Now we can find the velocity of the particle at $t=2\,\mathrm{s}$ by plugging in $t=2$:

$$ v(2\,\mathrm{s}) = 10(2) - 4\,\mathrm{m/s} = 16\,\mathrm{m/s} $$

Therefore, the magnitude of the velocity of the particle at $t=2\,\mathrm{s}$ is:

$$ \boxed{|v(2\,\mathrm{s})| = 16\,\mathrm{m/s}} $$