= Solution
In vacuum, with possible sources $\rho$ and $\mathbf J$, <Maxwell's equations> are
$$
\nabla\cdot\mathbf E=\rho/\epsilon_0,\quad\nabla\cdot\mathbf B=0,\quad\nabla\times\mathbf E=-\partial_t\mathbf B,\quad\nabla\times\mathbf B=\mu_0\mathbf J+\mu_0\epsilon_0\partial_t\mathbf E.
$$
Their integral forms, with compatible orientations and fixed surfaces, are
$$
\begin{aligned}
\oint_{\partial V}\mathbf E\cdot d\mathbf S&=\frac1{\epsilon_0}\int_V\rho\,dV,&\quad\oint_{\partial V}\mathbf B\cdot d\mathbf S&=0,\\
\oint_{\partial S}\mathbf E\cdot d\mathbf l&=-\frac d{dt}\int_S\mathbf B\cdot d\mathbf S,&\quad\oint_{\partial S}\mathbf B\cdot d\mathbf l&=\mu_0\int_S\mathbf J\cdot d\mathbf S+\mu_0\epsilon_0\frac d{dt}\int_S\mathbf E\cdot d\mathbf S.
\end{aligned}
$$
Charge- and current-free space additionally sets $\rho=0$, $\mathbf J=0$. Vacuum constitutive constants alone do not imply that no charges exist.
Take the <curl> of the source-free <Ampere-Maxwell law>, use <Faraday's law of induction> and $\nabla\cdot\mathbf B=0$:
$$
\nabla\times(\nabla\times\mathbf B)=-\nabla^2\mathbf B=-\mu_0\epsilon_0\partial_t^2\mathbf B.
$$
Consequently
$$
\boxed{\nabla^2\mathbf B-\frac1{c^2}\partial_t^2\mathbf B=0,\qquad c=\frac1{\sqrt{\mu_0\epsilon_0}}.}
$$
For the specified complex <plane wave>, the <wavevector> is $k\hat{\mathbf z}$ and $\omega=ck$. <Faraday's law of induction> fixes
$$
\boxed{\mathbf B=\frac{E_0}{c}e^{i(\omega t-kz)}\hat{\mathbf y}.}
$$
The physical fields are the real parts. They are in phase and their <Poynting vector> points along $+z$.
The instantaneous electric and magnetic <energy densities> are $u_E=\epsilon_0E^2/2$ and $u_B=B^2/(2\mu_0)$. Since $E=cB$, they are equal for this ideal travelling plane wave, and so are their cycle averages: \b[$\langle u_E\rangle/\langle u_B\rangle=1$]. For real peak amplitude $E_0$, each average is $\epsilon_0E_0^2/4$. The <Poynting vector> and mean intensity satisfy
$$
\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B,\qquad\boxed{\mathcal I=\langle S\rangle=\frac12\epsilon_0cE_0^2,\quad B_0=E_0/c.}
$$
It represents transported energy per area per time, in $\mathrm{W\,m^{-2}}$.
For the numerical moonlight estimate, adopt $L_\odot=3.828\times10^{26}\,\mathrm W$ and lunar mean radius $R_M=1.7374\times10^6\,\mathrm m$. These additional astronomical inputs give solar irradiance at the Moon of $\mathcal I_\odot=L_\odot/(4\pi d_\odot^2)\simeq1361\,\mathrm{W\,m^{-2}}$.<Sun Fact Sheet by National Aeronautics and Space Administration>[Bulk parameters: luminosity 382.8 in units of 10^24 J/s]{ref}<Moon Fact Sheet by National Aeronautics and Space Administration>[Bulk parameters: volumetric mean radius 1737.4 km and solar irradiance 1361 W/m^2]{ref}
Under the simple isotropic point-reflector approximation, the intercepted power is $\pi R_M^2\mathcal I_\odot$, of which fraction $A=0.12$ is redistributed over $4\pi$ steradians. Thus
$$
\mathcal I_M=\frac{A\mathcal I_\odot R_M^2}{4d_M^2}\simeq8.34\times10^{-4}\,\mathrm{W\,m^{-2}},\qquad\boxed{E_0\simeq0.793\,\mathrm{V\,m^{-1}},\quad B_0\simeq2.65\,\mathrm{nT}.}
$$
These are equivalent sinusoidal peak amplitudes; the corresponding <root mean square> fields are smaller by $\sqrt2$. Actual moonlight is broadband and incoherent.
A total reflected fraction alone does not specify the directional brightness of a full Moon. For a <Lambertian sphere at full phase>, with the same total reflectance $A$, integrating its radiance gives $\mathcal I_M=(2A/3)\mathcal I_\odot(R_M/d_M)^2$, instead giving $E_0\simeq1.30\,\mathrm{V\,m^{-1}}$ and $B_0\simeq4.32\,\mathrm{nT}$. If 0.12 is instead the <geometric albedo>, the full-phase formula is $\mathcal I_M=0.12\mathcal I_\odot(R_M/d_M)^2$, giving $E_0\simeq1.59\,\mathrm{V\,m^{-1}}$. NASA's table itself distinguishes <Bond albedo> from <geometric albedo>.<Moon Fact Sheet by National Aeronautics and Space Administration>[Bulk parameters list Bond albedo 0.11 and geometric albedo 0.12]{ref} The numerical estimate therefore needs an explicit reflection convention; the first boxed values use the stated isotropic approximation.
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