The ratio of the area within the electron orbits for the first excited state to the ground state for the hydrogen atom is

First excited state implies n =2, while ground state implies n =1.

If radius of orbit in ground state,  ${\mathrm{r}}_1={\mathrm{a}}_0$ then radius of orbit in first excited state

$=r_2={\mathrm{n}}^2{r}_1=(2{)}^2{r}_1=4r_1=4a_0$$\therefore {\mathrm{A}}_1=\pi {\mathrm{r}}_1^2=\pi {\mathrm{a}}_0^2\text{and }{\mathrm{A}}_2=\pi {\mathrm{r}}_2^2=\pi {\left(4{\mathrm{a}}_0\right)}^2=16\pi {\mathrm{a}}_0^2$

Hence  $({\mathrm{A}}_2/{\mathrm{A}}_1)=(16/1)$