Answer
$\rho = \frac{3~H^2}{8~\pi~G}$
Work Step by Step
We can find an expression for the mass $M$:
$M = \frac{4}{3}\pi~r^3~\rho$
We can find an expression for the minimum value of the required density:
$v = \sqrt{\frac{2GM}{r}}$
$v = \sqrt{\frac{(2G)(\frac{4}{3}\pi~r^3~\rho)}{r}}$
$v = \sqrt{(2G)(\frac{4}{3}\pi~r^2~\rho)}$
$\frac{v}{r} = \sqrt{(2G)(\frac{4}{3}\pi~\rho)}$
$H = \sqrt{(2G)(\frac{4}{3}\pi~\rho)}$
$H^2 = (2G)(\frac{4}{3}\pi~\rho)$
$\rho = \frac{3~H^2}{8~\pi~G}$
