Quick guess: For a finite field $k$ with non-trivial automorphism $\sigma$, build the skew polynomial ring $k[X, \sigma]$ and divide out the (two-sided) ideal $< X^2 >$.

For a finite field $k$ with non-trivial automorphism $\sigma$, take the skew polynomial ring $k[X, \sigma]$ (reminder: these are polynomials with coefficients on the left $\sum a_i X^i$ with relation $Xa = \sigma(a)X$) and set $R := k[X, \sigma] / < X^n >$ (i.e. divide out the (two-sided) ideal generated by $X^n$).

The only left as well as right ideals in this ring are the two-sided ones, namely, the ones generated by one of the $X^k$ for $0 \le k \le n$. The key fact to see this is that the units of $R$ are precisely the polynomials $\sum a_i \bar X^i$ with $a_0 \neq 0$ (to see this, use e.g. a geometric series argument). In case $n=2$, it is easily written down explicitly, $R$ can be seen as the set of $a + bX$ with $a, b \in k$ and multiplication given by

$(a+bX)(c+dX) = ac + (b\sigma(c) + ad) X$;
for $a \neq 0$, we have $(a + bX)^{-1} = a^{-1} - a^{-1} \sigma(a^{-1}) b X$.

@ Dimros, concerning your questions to Ralph: A matrix ring $M_n(k)$ over a finite (division) field $k$ (and a fortiori, any product of these) will never meet your conditions, as for $n =1$ it is commutative and for $n >1$ it has left ideals which are not two-sided. So by Artin-Wedderburn, any example will have non-zero radical and its semisimple quotient will be a product of fields. On the other hand, every ring embeds into any matrix ring over itself, so every example is a subring of matrix rings. Maybe you want some special kind of subring.