To solve for the angle of projection $\theta$, we will first establish the relationship between the variables given and then derive the formula using kinematics.

The time $T$ for one revolution at speed $v$ in a circle of radius $R$ is related to the circumference of the circle by the formula:

$$ v = \frac{2\pi R}{T} $$

When the particle is projected with the same speed $v$ at an angle $\theta$ to the horizontal, its vertical component of velocity is given by $v_y=v\sin\theta$.

The maximum height $H$ reached by the projectile can be found from the kinematic equation:

$$ H = \frac{v_y^2}{2g} = \frac{(v\sin\theta)^2}{2g} $$

We are given that the maximum height attained $H$ is equal to $4R$, so:

$$ 4R = \frac{(v\sin\theta)^2}{2g} $$

Substitute $v$ from the first equation into the second one:

$$ 4R = \frac{((\frac{2\pi R}{T})\sin\theta)^2}{2g} $$

$$ 4R = \frac{(2\pi R\sin\theta)^2}{2gT^2} $$

$$ 4R = \frac{4\pi^2 R^2 \sin^2\theta}{2gT^2} $$

$$ 2gT^2 = \pi^2 R \sin^2\theta $$

Now solve for $\sin\theta$:

$$ \sin\theta =\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2} $$

Finally, to solve for $\theta$, take the inverse sine of both sides:

$$ \theta = \sin^{-1}\left(\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2}\right) $$

Therefore, the correct answer is:

Option A

$$ \sin^{-1}\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2} $$