= Solution
The <coherent state> satisfies $\langle a\rangle=\alpha$, $\langle a^\dagger\rangle=\alpha^*$ and $\langle a^\dagger a\rangle=|\alpha|^2$. The <canonical commutation relation> adds the vacuum term $\langle aa^\dagger\rangle=|\alpha|^2+1$. Therefore
$$
\langle\hat E\rangle=E_0(\alpha+\alpha^*),\qquad
\langle\hat E^2\rangle=E_0^2[\alpha^2+(\alpha^*)^2+2|\alpha|^2+1],
$$
and the <coherent-state electric-field variance> is
$$
\boxed{(\Delta E)^2=\langle\hat E^2\rangle-\langle\hat E\rangle^2=E_0^2,\qquad \Delta E=E_0.}
$$
Here $E_0$ is taken positive as the field-amplitude scale. The <coherent state> displaces the mean field without changing this vacuum <optical quadrature> noise.
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