Solution:

Since the primes are $2$ away from each other, we can make them equal to $m-1,m+1,$ where $m$ is their average.

Then, $$(m+1)^3-(m-1)^3 = 31106 ,$$ making $$m^3+3m^2+3m+1$$$$-(m^3-3m^2+3m-1)$$$$= 6m^2+2 = 31106.$$

Therefore, $m^2= 5184,$ so $m=72.$

The primes are therefore $71,73.$ The least prime greater than both of those is $79,$ and its digit sum is $16.$

Thus, the answer is E .