Answer
$0.0955$
Work Step by Step
The probability that a given available state will be occupied by an electron is given by
$P(E)=\frac{1}{e^{\frac{E-E_F}{KT}}+1}$ (Occupancy probability) …............(1)
We have to find $P(E)$ for above the Fermi energy.
At $T=320\;K$ and for $E-E_F=0.0620\;eV$, the exponential term in Eq. (1) is
$e^{\frac{E-E_F}{KT}}=e^{\frac{0.0620}{8.62\times10^{-5}\times320}}=9.4657$
so $P(E)=\frac{1}{9.4657+1}=0.0955$
Therefore, the probability that a state above the Fermi energy will be occupied at $T=3200\;K$ is $0.0955$
