$\left( d \right)\,\,\bar x = \frac{{7 + 8 + 9 + 7 + 8 + 7 + \lambda  + 8}}{8} = 8$

           $ \Rightarrow \frac{{54 + \lambda }}{8} = 8 \Rightarrow \lambda  = 10$

Now variance $ = {\sigma ^2}$

$ = \frac{\begin{array}{l}
{\left( {7 – 8} \right)^2} + {\left( {8 – 8} \right)^2} + {\left( {9 – 8} \right)^2} + {\left( {7 – 8} \right)^2} + {\left( {8 – 8} \right)^2}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( {7 – 8} \right)^2} + {\left( {10 – 8} \right)^2} + {\left( {8 – 8} \right)^2}
\end{array}}{8}$

$ \Rightarrow {\sigma ^2} = \frac{{1 + 0 + 1 + 1 + 0 + 1 + 4 + 0}}{8} = \frac{8}{8} = 1$

Hence, the variance is $1$.