Starting from the field operator of a free Hermitean scalar field at time $t=0$, $$\phi(\mathbf{x},0)=\int \! \frac{d^3 k}{(2\pi)^3 2 \omega(\mathbf{k})}\left[e^{i \mathbf{k}\cdot \mathbf{x}}a(\mathbf{k})+e^{-i \mathbf{k} \cdot \mathbf{x}}a^\dagger(\mathbf{k})\right], \qquad \omega(\mathbf{k})=\sqrt{\mathbf{k}^2+m^2},$$ the commutation relations of the creation and annihilation operators are given by $$\left[ a(\mathbf{k}), a^\dagger(\mathbf{k}^\prime) \right]= (2 \pi)^3 2 \omega(\mathbf{k}) \delta^{(3)}(\mathbf{k}-\mathbf{k}^\prime), \qquad \left[a(\mathbf{k}),a(\mathbf{k}^\prime) \right]=0.$$ As $a(\mathbf{k}) |0\rangle=0$ implies $\langle 0 |A(a) |0\rangle=0$, the determination of the variance of the smeared operator $A(a)$ in the vacuum state boils down to the computation of $$\langle 0 | A(a)^2 |0\rangle= \pi^{-3}a^{-6}\int \! d^3x \, e^{-\mathbf{x}^2/a^2}\int \! d^3y \, e^{-\mathbf{y}^2/a^2}\langle 0 |\phi(\mathbf{x},0) \phi(\mathbf{y},0) | 0\rangle, $$where $$ \langle 0 | \phi(\mathbf{x}, 0) \phi(\mathbf{y}, 0) | 0\rangle = \int \!\frac{d^3k}{(2\pi)^3 2 \omega(\mathbf{k})} e^{i \mathbf{k}\cdot (\mathbf{x}-\mathbf{y})}. $$ Using the Gauss integral formula $$\int \! d^3x \, e^{-\mathbf{x}^2/a^2}e^{i \mathbf{k} \cdot \mathbf{x}} =\pi^{3/2} a^3 e^{-a^2 \mathbf{k}^2/2},$$$$\int \! d^3x \, e^{-\mathbf{x}^2/a^2}e^{i \mathbf{k} \cdot \mathbf{x}} =\pi^{3/2} a^3 e^{-a^2 \mathbf{k}^2/4},$$ the integration over the variables $\mathbf{x}$ and $\mathbf{y}$ can be performed, yielding $$\langle 0 |A(a)^2 |0\rangle=\frac{1}{4 \pi^2}\int\limits_0^\infty \frac{dk \, k^2}{\sqrt{k^2+m^2}}e^{-a^2k^2/2}=\frac{1}{4\pi^2 a^2}\int\limits_0^\infty\frac{ds s^2}{\sqrt{s^2+(ma)^2}} e^{-s^2/2}.$$ The result is dimensionally correct as $[\phi]=[A]=[m]=1$ and $[a]=-1$. It is now an easy task to work out the requested asymptotic expansions in the dimensionless variable $ma$ of the remaining integral.
