To solve this problem, we use the concept that for the coin to perfectly conceal the moon, it must subtend the same angle at the observer's eye as the moon's angular diameter. This geometry can be approximated by a sector of a circle.

1. Convert the angular diameter from arcminutes to radians:
The formula we will use requires the angle to be in radians. The given angular diameter is \(\theta = 31'\) (31 arcminutes).
First, we convert arcminutes to degrees, knowing that \(1^\circ = 60'\):
\[\theta = \left(\frac{31}{60}\right)^\circ\]
Next, we convert degrees to radians, knowing that \(180^\circ = \pi\) radians:
\[\theta = \frac{31}{60} \times \frac{\pi}{180} = \frac{31\pi}{10800} \text{ radians}\]
2. Use the arc length formula to find the distance:
For a small angle, the relationship between arc length (s), radius (r), and the angle in radians (\(\theta\)) is \(s \approx r\theta\).
In this problem's setup:
- The 'arc length' \(s\) is the diameter of the coin, \(d = 2\) cm.
- The 'radius' \(r\) is the distance from the eye to the coin, which we need to find. Let's call it \(D\).
Therefore, our formula is \(d = D \cdot \theta\).
We can rearrange this to solve for the distance \(D\):
\[D = \frac{d}{\theta}\]
3. Substitute values and calculate the distance:
Now we plug in the values for the coin's diameter and the angle in radians.
\[D = \frac{2 \text{ cm}}{\frac{31\pi}{10800}} = \frac{2 \times 10800}{31\pi} = \frac{21600}{31\pi} \text{ cm}\]
Using the approximation \(\pi \approx 3.14159\):
\[D \approx \frac{21600}{31 \times 3.14159} \approx \frac{21600}{97.389} \approx 221.78 \text{ cm}\]
4. Convert the distance to meters:
The answer options are given in meters, so we convert our result from centimeters to meters by dividing by 100.
\[D = \frac{221.78}{100} = 2.2178 \text{ m}\]
Result:
The coin should be held at a distance of approximately 2.217 m from the eye.