A

Step 1: Let the initial velocity of projection of the particle be $u$ and the angle of projection be $\theta$. The horizontal and vertical components of the initial velocity are:
$u_x = u \cos \theta$ and $u_y = u \sin \theta$.

Step 2: The equations of motion give:
For horizontal distance $a$:
$$h = a \tan \theta - \frac{g a^2}{2 u^2 \cos^2 \theta}$$ and

For horizontal distance $2a$:
$$h' = 2a \tan \theta - \frac{g (2a)^2}{2 u^2 \cos^2 \theta}$$

Here, $h'$ is the height at $2a$.

Step 3: The heights are equal (since both are defined as $h$), thus equating the two:
$$a \tan \theta - \frac{g a^2}{2 u^2 \cos^2 \theta} = 2a \tan \theta - \frac{g (4a^2)}{2 u^2 \cos^2 \theta}$$

Step 4: Rearranging the terms gives:
$$a \tan \theta - 2a \tan \theta = - \frac{g a^2}{2 u^2 \cos^2 \theta} + 2 \cdot \frac{g (4a^2)}{2 u^2 \cos^2 \theta}$$
$$( - a \tan \theta) = \frac{g a^2}{2 u^2 \cos^2 \theta} \cdot (8 - 1)$$

This simplifies to:
$$ a \tan \theta = \frac{7 g a^2}{2 u^2 \cos^2 \theta}$$

Step 5: Solve for $u^2$ gives:
$$ u^2 = \frac{7 g a}{2 \tan \theta \cos^2 \theta}$$
Using $\tan \theta = \frac{\sin \theta}{\cos \theta}$:
$$ u^2 = \frac{7 g a \cos^2 \theta}{2 \sin \theta \cos^2 \theta} = \frac{7 g a}{2 \sin \theta}$$

Thus, the velocity of projection is:
$$u = \sqrt{\frac{7 g a}{2 \sin \theta}}$$
Therefore, the required velocity of projection is given by this formula.