I want an example of entire function with order $\sqrt 2$.
If $f$ is a entire function of finite order $\rho$ then $\rho=\lim_{R \to \infty} \sup_{r \geq R} \frac{\log \log M(f,r)}{\log(r)}$, where $M(f,r)=\max_{|z|=r}|f(z)|$.
Take $f(z)=e^{z^{\sqrt 2}}$, then $M(f,r)=\max_{|z|=r}|e^{z^{\sqrt 2}}|=e^{r^{\sqrt 2}}$ for any $r>0$. Thus by the above formula $$\rho(f)=\sup_{r} \frac{\log \log (e^{r^{\sqrt 2}})}{\log r}=\sup_{r} \frac{\log (r^{\sqrt 2}\log e)}{\log r}=\sup_{r} \frac{\log (r^{\sqrt 2})}{\log r}=\sup_{r} \frac{\sqrt 2\log r}{\log r}=\sqrt 2.$$
So I think this serves as an example of entire function of order $\sqrt 2$.