= Solution
{c}
A <Cartesian closed category> has finite <products in a category> and, for every $F$, an <exponential object> $G^F$ characterized naturally by
$$
\mathcal E(K\times F,G)\cong\mathcal E(K,G^F).
$$
In $[\mathcal C,\mathbf{Set}]$, finite products and the terminal functor are computed pointwise. With $h_B=\mathcal C(B,-)$, define
$$
(G^F)(B)=\operatorname{Nat}(h_B\times F,G).
$$
This is a set because $\mathcal C$ is a <small category>. For $v:B\to B'$, precomposition $h_{B'}\to h_B$ induces $(G^F)(v)$ by precomposition of <natural transformations>, making $G^F$ a covariant <functor>.
Given $\alpha:K\times F\to G$, define its curry $\widehat\alpha:K\to G^F$ by
$$
\bigl(\widehat\alpha_B(y)\bigr)_C(u,x)=\alpha_C(K(u)y,x)
\qquad(u:B\to C,\ x\in F(C)).
$$
For $a:C\to C'$, naturality of $\alpha$ gives $G(a)\alpha_C(K(u)y,x)=\alpha_{C'}(K(au)y,F(a)x)$, proving that this is a <natural transformation> $h_B\times F\to G$. The same formula with $u$ precomposed proves naturality of $\widehat\alpha$ in $B$.
Conversely, from $\beta:K\to G^F$ define
$$
\alpha_B(y,x)=\bigl(\beta_B(y)\bigr)_B(1_B,x).
$$
Naturality of $\beta$ and its values proves naturality of $\alpha$. Currying this result recovers $\beta$, since $\beta_C(K(u)y)=(G^F)(u)(\beta_B(y))$; uncurrying a curry recovers $\alpha$ by setting $u=1_B$. These mutually inverse constructions are natural in $K,G$, giving the adjunction required for an <exponential object>. Therefore
$$
\boxed{[\mathcal C,\mathbf{Set}]\text{ is cartesian closed},\quad (G^F)(B)=\operatorname{Nat}(h_B\times F,G).}
$$
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