The normalised wave function of $2s$ orbital is $$\Psi_{2s} = \frac{1}{4 \sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{\frac{3}{2}}\left(2 - \frac{Zr}{a_0}\right)e^{-\frac{Zr}{2a_0}}$$ Now the radial probability distribution function will be $$P(r) = |\Psi|^2 4\pi r^2 = \frac{Z^3}{8a_0^3}r^2\left(2-\frac{Zr}{a_0}\right)^2 e^{-\frac{Zr}{a_0}},$$ so that $\int_{0}^{\infty} P(r) ~\mathrm dr =1.$ Now for getting the maxima and minima, solve $\frac{\mathrm d}{\mathrm dr}(P(r)) = 0.$ Simply we have to solve $$\frac{\mathrm d}{\mathrm dr}\left[r^2\left(2-\frac{Zr}{a_0}\right)^2 e^{-\frac{Zr}{a_0}}\right] =0.$$ Solving this one root will be from the equation $(2- \frac{Zr}{a_0})=0$, which will give you the value of $r$ for which $P(r) =0$, that will be the minima as you can easily verify. The maximas will come from the other quadratic equation $$\left(\frac{Zr}{a_0}\right)^2 -6\left(\frac{Zr}{a_0}\right) + 4 =0$$ which will give you two root as $\frac{Zr}{a_0} = 3+ \sqrt{5} ;\, 3-\sqrt{5}.$
So, the difference between two peak points will be $r_d =\frac{2\sqrt{5} a_0}{Z}.$ If it is Hydrogen atom, the distance will be $2\sqrt{5} \times0.529 A^0 = 2.365 A^0.$