Solution.
Let f be such a function, and let m≥0 be such that
g(z)=zmf(z)−z3
is still entire. Notice that g is still entire of second order and that
g(n1/3)=nm/3f(n1/3)−n=0
for all n∈N. Suppose that g is not identically zero, so that we may apply Jensen's formula:
log∣g(0)∣+∣an∣<R∑log∣an∣R=2π1∫02πlog∣∣g(Reiθ)∣∣dθ≲R2
where an denotes the zeroes of g. Then because each n1/3 is a root,
R2≳n1/3<R∑logn1/3R≳n<R3∑logR1/3R≳⌊R3⌋logR,
which is impossible. Hence, g must have been identically zero to begin with, i.e., f(z)=z3, so z3 is the only such function.