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Every recursively presented (even infinitely generated) group can be effectively embedded into a finitely generated recursively presented group either by using HNN extensions (as in Higman-Neumann-Neumann original paper) or by using small cancelation quotients of the free group. Every finitely generated recursively presented group can be effectively embedded into a finitely presented group by the Higman embedding theorem. The finite presentation is explicitly constructed using any Turing machine recognizing the set of relations of the finitely generated group. This means that one can explicitly write down a finite presentation of a group containing $\mathbb Q$. The number of generators and the number of relations will depend (linearly) on the number of commands in the Turing machine recognizing the defining relations of $\mathbb Q$. The real question is to find a "natural" finitely presented group containing $\mathbb Q$. That is not known so far. There are finitely presented groups containing close relatives of $\mathbb Q$. The Baumslag-Solitar group $BS(1,d)$ contains the group of $d$-adic rationals. And the R.Thompson group $V$ contains the group ${\mathbb Q}/{\mathbb Z}$.

Update 1. The first step can be simplified for $\mathbb Q$. For every $n\ge 2$ take the group $G_n=BS(1,n)=\langle a_n,b_n\mid b_n^{-1}a_nb_n=a_n^n\rangle$. The direct product $\Pi G_n$ contains $\mathbb Q$. Add two generators $t,s$ to the presentation of the direct product and all relations $t^{-1}a_it=a_{i+1}$, $s^{-1}b_is=b_{i+1}$ for all $i\ge 1$. That is a finitely generated (by $a_1,b_1, t,s$) recursively presented group containing $\mathbb Q$. The presentation of that group can be easily recognized by a Turing machine. Then the Higman construction gives a presentation of a finitely presented group containing $\mathbb Q$. The presentation will contain something like a 100 generators and 100 relations (I did not compute exact numbers).

Update 2. In Valiev, M. K. Universal group with twenty-one defining relations. Discrete Math. 17 (1977), no. 2, 207–213, Valiev constructed an explicit presentation with 21 defining relations of a group containing all finitely presented groups, hence containing $\mathbb Q$ (earlier a 26-relator example was constructed by Boone and Collins). The difference with the example in Update 1 is that it is hard to describe an embedding of $\mathbb Q$ in that group. That embedding is defined by the Turing machine describing a presentation of $\mathbb Q$.

Every recursively presented (even infinitely generated) group can be effectively embedded into a finitely generated recursively presented group either by using HNN extensions (as in Higman-Neumann-Neumann original paper) or by using small cancelation quotients of the free group. Every finitely generated recursively presented group can be effectively embedded into a finitely presented group by the Higman embedding theorem. The finite presentation is explicitly constructed using any Turing machine recognizing the set of relations of the finitely generated group. This means that one can explicitly write down a finite presentation of a group containing $\mathbb Q$. The number of generators and the number of relations will depend (linearly) on the number of commands in the Turing machine recognizing the defining relations of $\mathbb Q$. The real question is to find a "natural" finitely presented group containing $\mathbb Q$. That is not known so far. There are finitely presented groups containing close relatives of $\mathbb Q$. The Baumslag-Solitar group $BS(1,d)$ contains the group of $d$-adic rationals. And the R.Thompson group $V$ contains the group ${\mathbb Q}/{\mathbb Z}$.
Update. The first step can be simplified for $\mathbb Q$. For every $n\ge 2$ take the group $G_n=BS(1,n)=\langle a_n,b_n\mid b_n^{-1}a_nb_n=a_n^n\rangle$. The direct product $\Pi G_n$ contains $\mathbb Q$. Add two generators $t,s$ to the presentation of the direct product and all relations $t^{-1}a_it=a_{i+1}$, $s^{-1}b_is=b_{i+1}$ for all $i\ge 1$. That is a finitely generated (by $a_1,b_1, t,s$) recursively presented group containing $\mathbb Q$. The presentation of that group can be easily recognized by a Turing machine. Then the Higman construction gives a presentation of a finitely presented group containing $\mathbb Q$. The presentation will contain something like a 100 generators and 100 relations (I did not compute exact numbers).
Every recursively presented (even infinitely generated) group can be effectively embedded into a finitely generated recursively presented group either by using HNN extensions (as in Higman-Neumann-Neumann original paper) or by using small cancelation quotients of the free group. Every finitely generated recursively presented group can be effectively embedded into a finitely presented group by the Higman embedding theorem. The finite presentation is explicitly constructed using any Turing machine recognizing the set of relations of the finitely generated group. This means that one can explicitly write down a finite presentation of a group containing $\mathbb Q$. The number of generators and the number of relations will depend (linearly) on the number of commands in the Turing machine recognizing the defining relations of $\mathbb Q$. The real question is to find a "natural" finitely presented group containing $\mathbb Q$. That is not known so far. There are finitely presented groups containing close relatives of $\mathbb Q$. The Baumslag-Solitar group $BS(1,d)$ contains the group of $d$-adic rationals. And the R.Thompson group $V$ contains the group ${\mathbb Q}/{\mathbb Z}$.