Answer
It will be an acceptable absolute temperature scale.
$T(S)_{icepointofwater}=732.04S$
Work Step by Step
Knowing that all linear absolute temperature scales are constant multiples of each other. This scale will we an acceptable absolute temperature scale. For example the boiling temperature in $K$ is $373.15K$ and in Smith scale is $1000$ so:
$$1S=\frac{1000}{373.15}K=2.68K$$
As you can see $1S$ is equal to $2.68K$, just differentiated by a constant.
Then the ice point of water will be:
$$T(S)_{icepointofwater}=2.68*273.15=732.04S$$
