$\underline{\mathbf{Answer:D}}$
$\underline{\mathbf{Solution:}}$
$\begin{align}\textbf{Given:} \;\;\;\mathrm{x^2 = t > 0 }\\ \text{On integrating above equation, we get:} \\ \ \mathrm{g’(t) = t^{\frac{3}{2}}}\\ \mathrm{g(t) = \frac{2}{5}t^{\frac{5}{2}}} + \mathrm C \\ \textbf{Given} \; \mathrm g(1) = 1\\ \Rightarrow \mathrm C = \dfrac{3}{5} \\ \Rightarrow \mathrm{g(x^2) = \dfrac{2}{5}x^5 + \dfrac{3}{5} \\ \text{and,}\; g(2^2) = \dfrac{2}{5}2^5 + \dfrac{3}{5} \\= \dfrac{67}{5}} \end {align}$
$\therefore \mathbf D$ is the correct option.