Any ordered field $F$ has characteristic $0$, so it contains a copy of $\mathbb{Z}$; by the universal property of the quotient field, the ring monomorphism $\mathbb{Z}\to F$ lifts to a monomorphism $\mathbb{Q}\to F$. We can identify $\mathbb{Q}$ with its image, so it's not restrictive to assume that $\mathbb{Q}\subseteq F$.
It's not really difficult: if $m/n\in\mathbb{Q}$, then we send it to
$$
\frac{f(m)}{f(n)}\in F
$$
where $f\colon \mathbb{Z}\to F$ is the (unique) monomorphism. Is this a field homomorphism? Just a check.
Now we come to the order. First of all, positive integers are positive in $(F,\prec)$: if $n>0$, then
$$
n=\underbrace{1+1+\dots+1}_{\text{$n$ times}}
$$
and therefore $0\prec n$. Conversely, if $n<0$, then
$$
n=-(\,\underbrace{\,1+1\dots+1}_{\text{$-n$ times}}\,)
$$
and so $n\prec0$.
Any element of $\mathbb{Q}$ can be represented as $m/n$ with $n>0$, because $a/b=(-a)/(-b)$, where $a,b\in F$, $b\ne0$. So, let $0\prec m/n$ in the ordering of $F$, with $n>0$. Then, by the properties of ordered fields,
$$
0\prec n\cdot\frac{m}{n}=m
$$
and therefore $m>0$. So a rational which is positive in $(F,\prec)$ is also positive in the usual order. A rational which is negative in $(F,\prec)$ is the opposite of a positive rational (in both orders).
Let $\beta=\{x_1,x_2,x_3,...x_n\}$ be the basis we need to find $n$. Possible linear combinations $=7^2\times7^2...\times7^2=7^{2n}=7^6\implies n=3$
Best Answer
If $K_1$ and $K_2$ are algebraically closed fields of the same uncountable cardinality and of the same characteristic, then they also have the same transcendence degree over their prime field, and are therefore isomorphic.