The outcomes of the 10 rolls $\left(X_i\right)$ are independent. A key property of variance is that for independent variables, the variance of the sum is the sum of the variances:
$$
\operatorname{Var}(S)=\operatorname{Var}\left(X_1+X_2+\cdots+X_{10}\right)=\operatorname{Var}\left(X_1\right)+\operatorname{Var}\left(X_2\right)+\cdots+\operatorname{Var}\left(X_{10}\right)
$$
Since all rolls are from the same fair die, they all have the same variance (let's call it $\operatorname{Var}(X)$ ). Therefore, the equation simplifies to:
$$
\operatorname{Var}(S)=10 \times \operatorname{Var}(X)
$$
To find the variance of a single roll, we use the formula:
$$
\operatorname{Var}(X)=E\left[X^2\right]-(E[X])^2
$$
Find $E[X]$ (the expected value of a single roll)
$$
E[X]=(1+2+3+4+5+6) \times \frac{1}{6}=\frac{21}{6}=3.5
$$
Find $E\left[X^2\right]$ (the expected value of the square of a single roll)
$$
\begin{gathered}
E\left[X^2\right]=\left(1^2+2^2+3^2+4^2+5^2+6^2\right) \times \frac{1}{6} \\\\
E\left[X^2\right]=(1+4+9+16+25+36) \times \frac{1}{6}=\frac{91}{6} \\\\
\operatorname{Var}(X)=\frac{91}{6}-(3.5)^2 \\\\
\operatorname{Var}(X)=\frac{182}{12}-\frac{147}{12}=\frac{35}{12} \\\\
\operatorname{Var}(S)=10 \times \operatorname{Var}(X)=10 \times \frac{35}{12} \\\\
\operatorname{Var}(S)=\frac{350}{12}=\frac{175}{6}
\end{gathered}
$$
