Answer
$1\times10^{4}\;K$
Work Step by Step
The naturally occurring population ratio $\frac{N_x}{N_0}$ of the two states is due to thermal agitation of the gas atoms
$\frac{N_x}{N_0}=e^{-\frac{(E_x-E_0)}{kT}}$
where, $(E_x-E_0)$ is the energy separation between the two states.
Taking logarithm in both side
$\frac{(E_x-E_0)}{kT}=\ln\frac{N_0}{N_x}$
Substituting the given values
$\frac{3.2}{8.62\times10^{-5}\times T}=\ln\frac{2.5\times10^{15}}{6.1\times10^{13}}$
$T\approx1\times10^{4}\;K$
Therefore, the temperature of the star’s atmosphere at that altitude is $1\times10^{4}\;K$
