$\newcommand\R{\mathbb R}$ A convenient way to derive the dual problem from a primal one is by using the minimax duality 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.$$\inf_{v>2|R|}\Big(v+\int_R \frac{\max(0,b)^2}{4 (v-2|x|)}\,dx\Big). $$