$\newcommand\R{\mathbb R}$
A convenient way to derive the dual problem from a primal one is by using the [minimax duality][1] for the Lagrangian, which is given here by the formula  $$L(f,v):=\int_R\big[2|x|f(x)+b\sqrt{f(x)}\big]\,dx-v\Big(\int_Rf(x)\,dx-1\Big),$$
where $|x|:=\|x\|$ and $b:=\beta$. Clearly, 
$$\sup_{f\ge0}\inf_{v\in\R} L(f,v)
=\sup\Big\{\int_R\big[2|x|f(x)+b\sqrt{f(x)}\big]\,dx\colon f\ge0,\int_R f(x)\,dx=1\Big\},$$
which is value of the primal problem. 

The value of the dual problem is 
$$\inf_{v\in\R}\sup_{f\ge0} L(f,v)
=\inf_{v\in\R}\Big(v+\sup\Big\{\int_R\big[2|x|f(x)+b\sqrt{f(x)}-vf(x)\big]\,dx\colon f\ge0\Big\}\Big)
=\inf_{v\in\R}\Big(v+\int_R s(|x|,v)\,dx\Big),
$$
where 
$$s(a,v):=\sup\{2at+b\sqrt t-vt\colon t\ge0\}.$$
For any real $a>0$, it is easy to see that 
$$s(a,v)=\frac{b^2}{4 (v-2 a)}$$
if $b>0$ and $v>2a$, $s(a,v)=0$ if $b\le0$ and $v\ge2a$, and $s(a,v)=\infty$ otherwise; in particular, $s(a,v)=\infty$ if $v<2a$. Thus, with $|R|:=\max\{|x|\colon x\in R\}$, the value of the dual problem is 
$$\inf_{v>2|R|}\Big(v+\int_R \frac{b^2}{4 (v-2|x|)}\,dx\Big),
$$
as desired. 

  [1]: https://people.eecs.berkeley.edu/~elghaoui/Teaching/EE227A/lecture7.pdf