To find the length of the second's pendulum at a height $h = 2R$ from the surface of the Earth, we must first understand that the length of a second's pendulum, $L$, is related to the gravitational acceleration, $g$, and the period, $T$, by the formula: $$ T = 2\pi\sqrt{\frac{L}{g}} $$ Since we are talking about a second's pendulum, the period, $T$, is 2 seconds (since it takes one second for the pendulum to swing in one direction and another second to swing back), thus $T = 2 \text{ seconds}$.

Now let's find the gravitational acceleration at height $h = 2R$ where $R$ is the radius of the earth. The general formula for gravitational acceleration at a height $h$ above the surface is: $$ g_h = \frac{g}{{\left(1 + \frac{h}{R}\right)}^2} $$ Plugging $h = 2R$ into the formula, we get:

$$ g_h = \frac{g}{{(1 + \frac{2R}{R})}^2} $$

$$ g_h = \frac{g}{{(1 + 2)}^2} $$

$$ g_h = \frac{g}{3^2} = \frac{g}{9} $$

So the gravitational acceleration at height $h$ is one-ninth of the gravitational acceleration at the surface of the Earth. Given that $g = \pi^2 \text{ m/s}^2$, we get: $$ g_h = \frac{\pi^2}{9} \text{ m/s}^2 $$

Now knowing the gravitational acceleration at height $h$ and with the period $T$ of 2 seconds, we can rearrange the formula for the second's pendulum to solve for the length $L_h$:

$$ 2 = 2\pi\sqrt{\frac{L_h}{g_h}} $$

$$ 1 = \pi\sqrt{\frac{L_h}{g_h}} $$

$$ \frac{1}{\pi} = \sqrt{\frac{L_h}{g_h}} $$

Squaring both sides, we get:

$$ \frac{1}{\pi^2} = \frac{L_h}{g_h} $$

Multiplying both sides by $g_h$ gives us the length $L_h$:

$$ L_h = \frac{g_h}{\pi^2} $$

Substituting $g_h$ into the equation yields:

$$ L_h = \frac{\pi^2}{9\pi^2} $$

$$ L_h = \frac{1}{9} \text{ m} $$

Therefore, the length of the second's pendulum at a height $h = 2R$ from the surface of the Earth is $\frac{1}{9}$ meters. The correct answer is Option A.