I am looking for an example of a semisimple commutative Banach algebra which admits a non-semisimple quotient.
Attempt from the comments:
"I take $A$ to be the algebra of all continuously differentiable functions. Let $I$ to be ideal of all functions from $A$ such that $f(0)=f'(0)=0$. I want to prove that $A/I$ is a two-dimensional algebra which has a one-dimensional radical."
Take $A$ to be the Wiener algebra $A(\mathbb T)$, the algebra of all continuous functions on $\mathbb T$ with absolutely convergent Fourier series; in other words, $A(\mathbb T)$ is the image of $\ell_1(\mathbb Z)$ under the Fourier transform. The product of $A(\mathbb T)$ is the pointwise product and the norm is induced by $\ell_1(\mathbb Z)$, $$\Vert f\Vert_A=\sum_{n\in\mathbb Z} \vert\widehat f(n)\vert$$ It is well known that the characters of $A(\mathbb T)$ are the evaluations at points of $\mathbb T$. In particular, $A$ is semi-simple. Moreover, if $I$ is a closed ideal in $A$ with hull $h(I)=E\subset\mathbb T$, then the characters of $A/I$ are exactly the evaluations $\delta_s$ with $s\in E$. In particular, if $f\in A$ is identically $0$ on $E=h(I)$, then the spectral radius of $[f]_{A/I}$ is $0$.
For any closed set $E\subset \mathbb T$, denote by $I(E)$ the ideal of all $f\in A$ such that $f\equiv 0$ on $E$, and by $J(E)$ the ideal of all $f\in A$ such that $f\equiv 0$ on a neighbourhood of $E$. Since $A$ is a regular Banach algebra we have $h(\overline{J(E)})=E$. Moreover, by a famous result of P. Malliavin there exist $non$-$spectral$ $sets$ for $A(\mathbb T)$, i.e. closed sets $E\subset \mathbb T$ such that $\overline{J(E)}\neq I(E)$.
Now, if $E$ is a non-spectral set for $A$, then $A/\overline{J(E)}$ is not semi-simple just by definition: if you take any $f\in I(E)$ which is not in $\overline{J(E)}$, then $[f]$ is not $0$ in $A/\overline{J(E)}$ but has spectral radius $0$.