According to given data,
\(\begin{array}{ccccc}
\hline \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} & \boldsymbol{x}_{\boldsymbol{i}} \boldsymbol{f}_{\boldsymbol{i}} & \overline{\boldsymbol{x}}-\boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}}\left(\overline{\boldsymbol{x}}-\boldsymbol{x}_{\boldsymbol{i}}\right)^{\mathbf{2}} \\
\hline 1 & 1 & 1 & 6 & 36 \\
\hline 2 & 2 & 4 & 5 & 50 \\
\hline 3 & 3 & 9 & 4 & 48 \\
\hline 4 & 4 & 16 & 3 & 36 \\
\hline 5 & 5 & 25 & 2 & 20 \\
\hline 6 & 6 & 36 & 1 & 6 \\
\hline 7 & 7 & 49 & 0 & 0 \\
\hline 8 & 8 & 64 & 1 & 8 \\
\hline 9 & 9 & 81 & 2 & 36 \\
\hline 10 & 10 & 100 & 3 & 90 \\
\hline
\end{array}\)
\(\begin{aligned}
\because \quad N & =\Sigma f_i=55, \bar{x}=\frac{\Sigma x_i f_i}{N} \\
& =\frac{10 \times 11 \times 21}{6 \times 55}=7 \\
\therefore \quad \sigma^2 & =\frac{\Sigma f_i\left(\bar{x}-x_i\right)^2}{N} \\
& =\frac{36+50+48+36+20+6+8+36+90}{55} \\
& =\frac{330}{55}=6
\end{aligned}\)
Hence, option (d) is correct.