Do you know a finite unitary reversible ring that is not isomorphic to its opposite? And the minimal with that property? Do you know a finite unitary reversible ring that is not isomorphic to its opposite? And the minimal with that property?
The examples of rings not isomorphic to their opposite that I know of are not reversible.
 A: Applying the following example with $q=2$ gives a finite unital reversible ring $R$ with $2^6=64$ elements which is not isomorphic to its opposite. I do not know if this is the cardinality-wise smallest possible example.
Fix a prime power $q$ and let $F$ and $E$ denote finite fields with $q$ and $q^3$ elements, respectively. We assume $F\subseteq E$ and let $\sigma$ denote a generator of $\mathrm{Gal}(E/F)\cong C_3$. Let $E[X;\sigma]$ denote the twisted polynomial ring (a.k.a. Ore extension) in the variable $X$ over $E$. That is, the elements of $E[X;\sigma]$ are polynomials $\sum_{i=0}^t a_i X^i$ with coefficients in $E$, they are added in the usual way, but the product is determined by setting $X  a = \sigma(a)X$ for all $a\in E$. Let
$$
R=E[X;\sigma]/\langle X^2\rangle,
$$
where $\langle X^2\rangle$ is the ideal generated by $X^2$. Writing $x$ for the image of $X$ in $R$, elements of $R$ can be written uniquely as $a_0+a_1x$ with $a_0,a_1\in E$ and the product in $R$ is given by $(a_0+a_1x)(b_0+b_1x)=(a_0b_0)+(a_0b_1+a_1\sigma(b_0))x$.
We claim that $R$ is reversible and $R\ncong R^\mathrm{op}$.
To see that $R$ is reversible, note that its Jacobson radical $J:=Ex$  satisfies $J^2=0$ and $R/J\cong E$. That is, $R$ is local and it Jacobson radical squares to zero. All such rings are  reversible, because if $f,g\in R$ satisfy $fg=0$, then $f,g$ must both lie in the radical and therefore satisfy $gf=0$.
Next, for the sake of contradiction, suppose that there is an anti-automorphism $\tau:R\to R$. Then $\tau(J)=J$, and thus $\tau$ induces an automorphism $\tau_0$ of $E\cong R/J$. Consequently, $\tau(a)-\tau_0(a)\in J$ for all $a\in E$ (we regard $E$ as a subring of $R$). Write $\tau(x)=ux$ with $u\in E-\{0\}$. Then for all $a\in E$, we have
$$
\tau(\sigma(a)x)=\tau(x)\tau(\sigma(a))
=ux\cdot \tau(\sigma(a))
=ux\cdot \tau_0(\sigma(a))=u\cdot \sigma\tau_0\sigma(a)\cdot x
$$
where the third equality holds because $\tau(\sigma(a))-\tau_0(\sigma(a))\in J$ and $xJ=0$. Likewise, one finds that
$$
\tau(xa)=\tau(a)\tau(x)=\tau_0(a)ux=u\cdot \tau_0(a)\cdot x.
$$
Since $xa=\sigma(a)x$ and $u\neq 0$, it follows that
$$\tau_0(a)=\sigma\tau_0\sigma(a)$$ for all $a\in E$, or rather
$\tau_0=\sigma\tau_0\sigma$ in $\mathrm{Gal}(E/F)$. Since $\mathrm{Gal}(E/F)\cong C_3$, this means that $\sigma^2=1$, which is absurd by our choice of $\sigma$.
