An atom $$^8_{3}X$$ is bombarded by shower of fundamental particles and in 10s this atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the smface area of the nucleons is recorded by:

The nucleus of an atom is usually treated as a sphere whose radius varies with the mass number $$A$$ according to the empirical relation $$R = R_0\,A^{1/3}$$, where $$R_0$$ is a constant.

For a sphere, surface area $$S$$ is $$4\pi R^2$$. Hence,

$$S \propto R^2 \;\Longrightarrow\; S \propto \left(A^{1/3}\right)^{2}=A^{2/3} \quad -(1)$$

Initial nucleus: $$^{8}_{3}X$$  ⇒  initial mass number $$A_i = 8$$.

During the $$10\,$$s bombardment the nucleus absorbs
 • $$10$$ protons  ($$p$$)
 • $$9$$ neutrons  ($$n$$)
(The captured electrons do not stay in the nucleus, so they do not alter the nucleon count.)
Total extra nucleons $$= 10 + 9 = 19$$.

Final mass number:
$$A_f = A_i + 19 = 8 + 19 = 27$$.

Using relation $$-(1)$$, the ratio of final to initial surface areas is

$$\dfrac{S_f}{S_i}= \left(\dfrac{A_f}{A_i}\right)^{2/3} =\left(\dfrac{27}{8}\right)^{2/3}$$

Evaluate the exponent step by step:
$$\dfrac{27}{8}=3.375 \quad\Longrightarrow\quad \left(3.375\right)^{1/3}=1.5$$ (because $$(1.5)^3 = 3.375$$).
Therefore,
$$\left(\dfrac{27}{8}\right)^{2/3}=(1.5)^2=2.25.$$(Final surface area is $$2.25$$ times the initial.)

Percentage growth is defined as

$$\text{Percentage growth}=\left(\dfrac{S_f}{S_i}\right)\times100 =2.25 \times 100 = 225\%.$$(This includes the initial 100%; the surface area becomes 225% of its original value.)

Hence, the percentage growth recorded is 225%.

Answer (Option A)