Here m=6 and k=8
Probability of slot not hashed = $\frac{m-1}{m}$
Probability that particular slot being empty after k insertion =$\begin{pmatrix} \frac{m-1}{m} \end{pmatrix}^{k}$
So not being empty for particular slot=$1- \begin{pmatrix} \frac{m-1}{m} \end{pmatrix}^{k}$
Each slot having same probability so= $m \times \begin{pmatrix} 1- \begin{pmatrix} \frac{m-1}{m} \end{pmatrix}^{k} \end{pmatrix}$
So ans should be: $6 \times \begin{pmatrix} 1- \begin{pmatrix} \frac{5}{6} \end{pmatrix}^{8} \end{pmatrix}$ = 4.6045