Suspended from a nail in a wall, a closed circular wire experiences...

The moment of inertia of the closed circular wire (ring) about the point of suspension on its circumference is $I = I_{cm} + M R^2 = 2 M R^2$.

The distance of the center of mass from the point of suspension is $d = R$.

The time period of the physical pendulum is given by:

$T = 2 \pi \sqrt{\dfrac{I}{M g d}} = 2 \pi \sqrt{\dfrac{2 M R^2}{M g R}} = 2 \pi \sqrt{\dfrac{2 R}{g}}$

Given $T = 2 \text{ s}$ and $g = \pi^2 \text{ m/s}^2$, we have:

$2 = 2 \pi \sqrt{\dfrac{2 R}{\pi^2}} \Rightarrow 1 = \sqrt{2 R} \Rightarrow 2 R = 1 \text{ m}$

The distance of the farthest point on the wire from the point of suspension is $L = 2 R = 1 \text{ m}$.

The amplitude of oscillation is \theta_0 = 2^{}^{{}^{\circ}} = 2 \times \dfrac{\pi}{180} = \dfrac{\pi}{90} \text{ rad}.

The angular frequency of the pendulum is $\omega = \dfrac{2 \pi}{T} = \pi \text{ rad/s}$.

The maximum angular velocity of the pendulum as it passes through the mean position is $\Omega = \theta_0 \omega = \dfrac{\pi^2}{90} \text{ rad/s}$.

At the mean position, the angular acceleration is zero, so the tangential acceleration is zero. The acceleration of the farthest point is purely centripetal:

$a = \Omega^2 L = \left( \dfrac{\pi^2}{90} \right)^2 \times 1 = \dfrac{\pi^4}{8100} \text{ m/s}^2$

Converting to $\text{cm/s}^2$:

$a = \dfrac{\pi^4}{81} \text{ cm/s}^2$

Using $\pi^2 \approx 9.87$, we get $\pi^4 \approx 97.4$.

$a \approx \dfrac{97.4}{81} \approx 1.2 \text{ cm/s}^2$

The direction of this centripetal acceleration is towards the center of its circular path, which is the point of suspension.

Answer: $1.2 \text{ cm/s}^2$ towards the point of suspension