The only possible meaning for smoothness that comes to mind is the following: A measure on ${\mathbb R}^n$ is smooth, if it has a smooth density against the Lebesgue-measure. On a manifold, a measure is smooth if it transforms to smooth measures on every smooth chart. The question, whether a given smooth measure comes from a metric is equivalent to the question whether its density has a zero or not. If it has no zero, simply choose any Riemannian metric. Then your given measure has a nowhere vanishing density against the measure coming from the metric. Simply multiply the metric with the reciprocal square-root a suitable power of the density to get a metric that induces the given measure. Since Radon-Nikodym densities are uniquely determined, this is an if and only if criterion.
The only possible meaning for smoothness that comes to mind is the following: A measure on ${\mathbb R}^n$ is smooth, if it has a smooth density against the Lebesgue-measure. On a manifold, a measure is smooth if it transforms to smooth measures on every smooth chart. The question, whether a given smooth measure comes from a metric is equivalent to the question whether its density has a zero or not. If it has no zero, simply choose any Riemannian metric. Then your given measure has a nowhere vanishing density against the measure coming from the metric. Simply multiply the metric with the reciprocal square-root of the density to get a metric that induces the given measure. Since Radon-Nikodym densities are uniquely determined, this is an if and only if criterion.