1. For equilibrium, the potential energy must be minimum, so $\frac{dU}{dr}=0.$

Given $U=\frac{\alpha}{r^{10}}-\frac{\beta}{r^5}-3.$

2. Differentiate with respect to $r$: $\frac{dU}{dr}=\alpha\frac{d}{dr}(r^{-10})-\beta\frac{d}{dr}(r^{-5})$ $\frac{dU}{dr}=-10\alpha r^{-11}+5\beta r^{-6}.$

So equilibrium condition is $-10\alpha r^{-11}+5\beta r^{-6}=0.$

3. Simplify: $5r^{-11}(-2\alpha+\beta r^5)=0.$ Since $r\neq 0$, we get $-2\alpha+\beta r^5=0$ $\beta r^5=2\alpha$ $r^5=\frac{2\alpha}{\beta}.$

Hence, $r=\left(\frac{2\alpha}{\beta}\right)^{1/5}.$

4. Comparing with the given form $\left(\frac{2\alpha}{\beta}\right)^{a/b},$ we get $\frac{a}{b}=\frac{1}{5}.$ Therefore, $a=1.$

5. Check for minimum: $\frac{d^2U}{dr^2}=110\alpha r^{-12}-30\beta r^{-7}.$ At equilibrium, using $\beta r^5=2\alpha$, $\frac{d^2U}{dr^2}=110\alpha r^{-12}-30(2\alpha)r^{-12}=50\alpha r^{-12}>0,$ so it is indeed stable equilibrium.