Since $|\mathcal{S}| = 3$ the easiest way will be by direct enumeration:
\begin{align*}
\mathbf{E}[S^T A S] & = \frac{1}{3} \bigg( (\delta,0,0)^T A (\delta, 0,0) +
(0,\delta,0)^T A (0,\delta, 0) +
(0,0,\delta)^T A (0,0,\delta) \bigg)
\end{align*}
If $A = (a_{i,j})$ with $1 \leq i \leq 3, 1 \leq j \leq3$ then
\begin{align*}
(\delta,0,0)^T A (\delta, 0,0) & =a_{1,1}\delta^2 \\
(0,\delta,0)^T A ( 0,\delta, 0) & =a_{2,2}\delta^2 \\
(0,0,\delta)^T A ( 0,0,\delta) & =a_{3,3}\delta^2
\end{align*}
Hence
$$
\mathbf{E}[S^T A S] = \delta^2(a_{1,1} + a_{2,2} + a_{3,3}) = \delta^2 \text{Tr}(A)
$$
If you are really keen to solve the problem using the covariance matrix, then let us denote $S = (s_1,s_2,s_3)$ for the random element of $\mathcal{S}$ and then the covariance matrix $\Sigma$ is given by
\begin{align*}
\Sigma_{i,j} \, \colon & = \mathbf{E}[S_iS_j] - \mathbf{E}[S_i]\mathbf{E}[S_j] \\
& = \delta^2\mathbf{1}_{i,j} - \mathbf{E}[S_i]\mathbf{E}[S_j] \\
& = \delta^2\mathbf{1}_{i,j} - \left(\frac{\delta}{3} \times \frac{\delta}{3} \right) \\
& =\delta^2\left( \mathbf{1}_{i,j} - \frac{1}{9}\right)
\end{align*}
where $\mathbf{1}_{i,j} = 1$ if $i = j$ and is $0$ otherwise. That $\mathbf{E}[S_iS_j] =\mathbf{1}{i,j}$ follows since only one element of $S$ is non-zero at any one time. Hence we have
$$ \Sigma = \delta^2 \left( I - \frac{1}{9} J \right)$$
where $I$ is the $3 \times 3$ identity matrix, and $J$ is the $3 \times 3$ matrix with all entries equal to 1.
