We know the Fermi distribution function,

$f(e)=\dfrac{1}{1+e^{\frac{E-E_F}{KT}}}$

Rearranging the above equation we get,

$e^{\frac{E-E_F}{KT}}=\dfrac{1}{f(e)}-1$

Taking the natural logarithm on both the sides of the equation we get,

$\dfrac{E-E_F}{KT}=ln\left(\dfrac{1}{f(e)}-1\right)$

Simplifying the above equation we get,

$E=E_F+KTln\left(\dfrac{1}{f(e)}-1\right)$

The first term in the above equation is in eV, while the second term is in joule, to convert the second term in to eV we divide it by the charge of electrons$1.6\times10^{-19}$

Substituting the values in above equation for $f(E)=0.99$

$E=2.1+\dfrac{1.38\times10^{-23}\times300}{1.6\times10^{-19}}ln\left(\dfrac{1}{0.99}-1\right)$

Therefore energy for the probability of occupying 0.09 is 1.98 eV.

Similarly,

Substituting the values in above equation for $f(E)=0.01$

$E=2.1+\dfrac{1.38\times10^{-23}\times300}{1.6\times10^{-19}}ln\left(\dfrac{1}{0.01}-1\right)$

Therefore energy for the probability of occupying 0.01 is 2.219 eV.