No Hilbert space is a $C^*$-algebra in any natural way since the structure of Hilbert spaces (a vector space with an inner

product) is quite different from that of a C*-algebra (a Banach space equipped with an involutive algebra structure).

Nevertheless from the data provided by the OP one may give $A$ the structure of a $C^*$-algebra in a very natural

way. The first step is to consider the hermitian form $g$ as an inner product on $A$, so that

$$

H:= (A,g)

$$

becomes a Hilbert space

(completeness follows from finite dimensionality).

One may then define a map

$$

pi :Ato B(H),

$$

by

$$

pi (a)xi = axi , quadforall ain A,quad forall xi in H,

$$

(here $axi $ is nothing other than the product of $a$ and $xi $).

One may easily show that $pi $ is a *-homomorphism. It is moreover injective since, for $aneq 0$, one has that

$pi (a)a^*=aa^*neq 0$, as a consequence of $varphi (aa^*)>0$.

Finally, identifying $A$ with its image within $B(H)$ via $pi $, we have that $A$ becomes a $C^*$-algebra.

**Later edit**:

The late edit indeed makes the question trivial and, as noted my Martin, the only example is the complex numbers.

A perhaps more elementary proof of this fact is that every finite dimensional $C^*$-algebra is the direct sum of matrix algebras, so unless $text{dim}(A)=1$, there exists a pair of mutually orthogonal projections $p$ and $q$. It follows that $|ppm q|=1$, hence the parallelogram law fails.

The following statements are equivalent:

$A$ is a C$^*$-algebra with the norm induced by $varphi$

$dim A=1$, i.e., $A=mathbb C$

*Proof.* if $A=mathbb C$, then the only state is the identity, which induces the norm.

Conversely, if $A$ is a C$^*$-algebra, its norm satisfies the C$^*$-identity

$$tag1|x|^2=|x^*x|.$$ In terms of $varphi$, this is

$$tag2varphi(x^*x)=varphi((x^*x)^2)^{1/2},qquad xin A.$$

So, for each positive $ain A$,

$$tag3varphi(a)^2=varphi(a^2).$$

As $varphi$ is a ucp map, the equality $(3)$ says that $a$ is in the multiplicative domain of $varphi$. Thus the multiplicative domain of $varphi$ is all of $A$, since a C$^*$-algebra is spanned by its positive elements. So $varphi$ is a faithful representation of $A$ into (thus onto) $mathbb C$.