A

Step 1: Understand the basics of circular motion. In uniform circular motion, the particle is moving with a constant speed but changing direction continuously. Therefore, it experiences acceleration known as centripetal acceleration.
Step 2: For a particle moving in a circle of radius $r$ with speed $v$, the centripetal acceleration $a_c$ is given by the formula:
$$ a_c = \frac{v^2}{r} $$
Substituting the known values, where $v = 5 \, \text{ms}^{-1}$ and $r = 5 \, \text{m}$:
$$ a_c = \frac{(5)^2}{5} = \frac{25}{5} = 5 \, \text{ms}^{-2} $$
Step 3: However, we need the average acceleration during half a revolution. In half a revolution, the particle moves from one side of the circle to the opposite side. The change in velocity is significant here.
Step 4: The initial velocity $\vec{v_i}$ is along the tangent to the circle at one point, and the final velocity $\vec{v_f}$ is along the tangent at the opposite point. Thus,
$$ \vec{v_i} = 5 \, \hat{i} \; \text{(at initial point)} \text{ and } \vec{v_f} = -5 \, \hat{i} \; \text{(at opposite point)} $$
Step 5: The change in velocity $\Delta \vec{v}$ is:
$$ \Delta \vec{v} = \vec{v_f} - \vec{v_i} = -5 \, \hat{i} - 5 \, \hat{i} = -10 \, \hat{i} $$
Step 6: The average acceleration $\bar{a}$ over time $t$ is given by:
$$ \bar{a} = \frac{\Delta \vec{v}}{\Delta t} $$
The time $\Delta t$ for half a revolution in circular motion (circumference $C = 2\pi r = 10\pi$ m and speed = 5 m/s) is:
$$ \Delta t = \frac{10\pi}{5} = 2\pi \, \text{s} $$
Step 7: Therefore, the average acceleration becomes:
$$ \bar{a} = \frac{-10}{2\pi} = -\frac{5}{\pi} \, \text{ms}^{-2} $$
Step 8: The magnitude of the average acceleration is:
$$ |\bar{a}| = \frac{5}{\pi} \, \text{ms}^{-2} $$
Conclusion: The magnitude of average acceleration is approximately $1.59 \, \text{ms}^{-2}$. Hence, the correct answer option is A.