= Solution
A general-purpose high-dimensional choice is <Monte Carlo integration>. Draw <independent> <vectors> $W_r$ uniformly from the unit cube and evaluate $H(W_r)=\sum_{i=1}^{100}e^{\|W_r-a_i\|^2}$. The cube has volume $1$, so its <integral> is the <expected value> of $H(W_1)$. The iterative estimator is
$$
y_N=\frac1N\sum_{r=1}^NH(W_r),\qquad
y_{N+1}=y_N+\frac{H(W_{N+1})-y_N}{N+1}.
$$
The integrand is <continuous> on a compact <set>, hence bounded with finite <variance>. The <strong law of large numbers> gives almost sure convergence, and the square root of the <mean squared error> is exactly $\sigma/\sqrt N$, where $\sigma^2=\operatorname{Var}(H(W_1))$. Thus the standard stochastic accuracy rate is \b[$O(N^{-1/2})$]; the dimension affects the <variance> and evaluation cost, not this exponent.
There is also an important structure-specific improvement here: $e^{\|x-a_i\|^2}=\prod_{j=1}^{100}e^{(x_j-a_{ij})^2}$ is a <separable integrand>. The <Fubini theorem> reduces the requested value exactly to
$$
y=\sum_{i=1}^{100}\prod_{j=1}^{100}\int_0^1e^{(t-a_{ij})^2}\,dt.
$$
One can therefore replace high-dimensional sampling by $10000$ one-dimensional <composite Simpson's rule> calculations. For each factor use even $n$, $h=1/n$, and
$$
Q_n(g)=\frac h3\left[g(0)+g(1)+4\sum_{r=1,\ r\text{ odd}}^{n-1}g(rh)+2\sum_{r=2,\ r\text{ even}}^{n-2}g(rh)\right],
$$
then iterate $n\leftarrow2n$ and form the sum of the products. These smooth factors give \b[$O(n^{-4})$ deterministic discretization error] for fixed centres. Thus <Monte Carlo integration> is a valid generic answer, while exploiting separability is the more efficient specialized alternative; a large dimension alone does not force a tensor-product grid.
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