We know that delta function has the amazing property of
$\int_{a}^{b} δ(x-X) f(x) $ = $f(X)$
So, can it be carried over to three dimensions?
$$\int_{a}^{b} \int_{c}^{d} \int_{e}^{f} δ(x-X)δ(y-Y)δ(z-Z) f(x)f(y)f(z) dx dy dz = f(X)f(Y)f(Z)$$
Where,. $a<X<b, c<Y<d, e<Z<f$ are regions that define the volume where the Delta function is located.
Actually I disagree with your defining property of the $\delta$ function - what is $X$? What is the function $f$? The one dimensional $\delta$ function you're trying to define is $$\int_{a}^{b} \,dx \delta(x - X) f(x) = f(X)$$ as long as $X \in (a, b)$.
In fact we can do better: we define a three-dimensional delta function as follows: $$\delta^{3}(\mathbf{x} - \mathbf{X}) = \delta(x_{1} - X_{1})\delta(x_{2} - X_{2})\delta(x_{3} - X_{3})$$ which has the property that for well-behaved functions $f(\mathbf{x}): \mathbb{R}^{3} \rightarrow \mathbb{R}$, say, $$\int d^{3} x\, \delta^{3}(\mathbf{x} - \mathbf{X}) f(\mathbf{x}) = f(\mathbf{X})$$ asumming the point $\mathbf{X}$ lies within the integration region.