= Solution
Write the <Cartesian comonad> as $(G,\varepsilon,\delta)$, with $G$ preserving <finite limits>. A <coalgebra for a comonad> is $(A,a)$ with $a:A\to GA$, $\varepsilon_Aa=1_A$ and $\delta_Aa=Ga\,a$. Its <forgetful functor> $U:\mathcal E^G\to\mathcal E$ is faithful, creates <finite limits>, and has the <cofree coalgebra> right adjoint $R(X)=(GX,\delta_X)$. We construct the two remaining <topos> structures explicitly.
For <exponential objects>, take coalgebras $(A,a),(B,b)$ and start with $R(B^A)$. Put $X=G(B^A)$, with structure $\xi=\delta_{B^A}$, and let
$$
e:X\times A\longrightarrow B,\qquad e=\operatorname{ev}(\varepsilon_{B^A}\times1_A).
$$
Transpose the following two maps $X\times A\to GB$ into maps $u,v:X\to(GB)^A$:
$$
be,\qquad Ge\circ(\xi\times a),
$$
using $G(X\times A)\cong GX\times GA$. The cofree <adjunction> transposes $u,v$ once more into coalgebra morphisms $R(B^A)\rightrightarrows R((GB)^A)$. Let $Q$ be their <equalizer>. A coalgebra map $(C,c)\to R(B^A)$ corresponds to an arbitrary ambient map $h:C\times A\to B$. It factors through $Q$ precisely when
$$
bh=Gh\circ(c\times a),
$$
which is exactly the condition that $h$ be a coalgebra morphism. Therefore $Q$ represents $\mathcal E^G(C\times A,B)$ and is the required <exponential in a coalgebra topos>.
For the <subobject classifier>, let $\kappa:G\Omega\to\Omega$ classify the mono $G\top:G1\cong1\hookrightarrow G\Omega$. In the cofree coalgebra $R\Omega$, form
$$
\Omega_G=\operatorname{Eq}\bigl(1_{R\Omega},G\kappa\,\delta_\Omega\bigr).
$$
Both arrows are coalgebra morphisms. The cofree transpose of $\top:1\to\Omega$ factors through this <equalizer> and gives its true arrow. To verify classification, let $S\hookrightarrow X$ have ambient characteristic map $\chi:X\to\Omega$. It supports a subcoalgebra of $(X,x)$ exactly when it is invariant under $x$, equivalently
$$
S=x^{-1}(GS),\qquad \chi=\kappa G\chi\,x.
$$
The counit proves the reverse containment in the first equation; the forward containment supplies the restricted structure map, whose coalgebra laws follow through the mono. Under the cofree <adjunction>, the second equation says exactly that $G\chi\,x:X\to R\Omega$ factors through $\Omega_G$. Pulling back its true arrow recovers $S$, since $\kappa G\chi\,x=\chi$. This proves the universal property of the <subobject classifier of a coalgebra topos>. Hence \b[$\mathcal E^G$ is a <topos>].
Now let $f:\mathcal E\to\mathcal F$ be a <geometric morphism>, with $L=f^*\dashv H=f_*$. The comonad $G=LH$ is Cartesian: $L$ preserves <finite limits> and the right adjoint $H$ preserves limits. Put $\mathcal D=\mathcal E^G$, already a <topos>. The forgetful <adjunction> $U\dashv R$ defines $p:\mathcal E\to\mathcal D$ with $p^*=U$, so \b[$p^*$ is faithful].
The comparison <functor>
$$
K:\mathcal F\to\mathcal D,\qquad K(Y)=(LY,L\eta_Y)
$$
preserves <finite limits>. It has a right adjoint $J$, given on a coalgebra $(A,a)$ by
$$
J(A,a)=\operatorname{Eq}\bigl(Ha,\eta_{HA}:HA\rightrightarrows HGA\bigr).
$$
Indeed, the transposed arrow $Y\to HA$ corresponds to a coalgebra map $KY\to(A,a)$ exactly when it equalizes these two maps. Applying the finite-limit-preserving $L$ shows that $LJ(A,a)$ is the <equalizer> of
$$
Ga,\delta_A:GA\rightrightarrows G^2A.
$$
That <equalizer> is $a:A\to GA$. If $t:Z\to GA$ equalizes the pair, then $a\varepsilon_At=\varepsilon_{GA}Ga\,t=\varepsilon_{GA}\delta_At=t$, proving the claimed universal property. Consequently the counit $KJ(A,a)\to(A,a)$ is invertible. The <fully faithful adjoint criterion> makes \b[$J$ full and faithful].
Thus $K\dashv J$ defines a <geometric embedding> $i:\mathcal D\to\mathcal F$, and $UK=L$ identifies the composite with $f$. The requested factorization is
$$
\boxed{\mathcal E\xrightarrow{\ p\ }\mathcal D\xrightarrow{\ i\ }\mathcal F,\qquad f=i\circ p,}
$$
where $p$ is a <surjective geometric morphism> and $i_*=J$ is full and faithful.
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