Occupation probability 3kT above Fermi level

The probability that a state of energy $E$ is occupied by an electron is given by the Fermi–Dirac distribution function:

$$f(E) = \frac{1}{1 + \exp\left(\dfrac{E - E_F}{kT}\right)}$$

Given: the state is $3kT$ above the Fermi level, so $E - E_F = 3kT$.

$$\begin{aligned} f(E) &= \frac{1}{1 + e^{3kT/kT}} = \frac{1}{1 + e^{3}} \\ &= \frac{1}{1 + 20.09} = \frac{1}{21.09} = 0.0474 \end{aligned}$$

Answer: $f(E) = 0.0474$, i.e. about 4.74 %.

The result does not depend on temperature, because the energy is given in units of $kT$. A state only $3kT$ above $E_F$ (about 0.078 eV at 300 K) is already rarely occupied; the probability that it is empty is $1 - 0.0474 = 0.9526$.