This is tough may be not for u

$E$ occurs when the selected number is one of $\{29, 92, 36, 63\}$ .

Probability that $E$ occurs $1$ times for $1$ number $= \frac{1}{25}$

Probability that $E$ occurs $3$ times for $4$ numbers $= 4 \times \frac{1}{25} \times \frac{1}{25} \times \frac{1}{25} \times \frac{24}{25} = \frac{96}{25^4}$

Probability that $E$ occurs $4$ times for $4$ numbers $= \frac{1}{25} \times \frac{1}{25} \times \frac{1}{25} \times \frac{1}{25} = \frac{1}{25^4}$

Probability that $E$ occurs at least $3$ times $= \frac{96}{25^4} + \frac{1}{25^4} = \frac{97}{25^4}$

$\Rightarrow k = 97$