The distance covered by a body in the $$n^{\text{th}}$$ second can be found using the equation:

$$S_{n} = u + \dfrac{1}{2}a(2n-1)$$

where,

• $$S_{n}$$ is the distance covered in the $$n^{\text{th}}$$ second,
• $$u$$ is the initial velocity,
• $$a$$ is the acceleration, and
• $$n$$ is the nth second.

The distance covered in the $$n^{\text{th}}$$ second is given as $$102.5 \, \text{m}$$, so we have:

$$102.5 = u + \dfrac{1}{2}a(2n-1)$$ ---- (1)

For the $$(n + 2)^{\text{th}}$$ second, the distance covered is:

$$S_{n+2} = u + \dfrac{1}{2}a(2(n+2)-1)$$

Substituting $$n + 2$$ in place of $$n$$, we get:

$$115.0 = u + \dfrac{1}{2}a(2n+3)$$ ---- (2)

Subtracting equation (1) from equation (2), we get:

$$115.0 - 102.5 = \dfrac{1}{2}a(2n + 3 - 2n + 1)$$

$$12.5 = \dfrac{1}{2}a(4)$$

So, solving for $$a$$ gives:

$$a = \dfrac{12.5 \times 2}{4} = \dfrac{25}{4} = 6.25 \, \text{m/s}^2$$

Therefore, the acceleration of the body is:

$$6.25 \, \text{m/s}^2$$

Which corresponds to Option A: $$6.25 \, \text{m/s}^2$$.