Mean Value Property Implies "Volume" Mean Value Property

Suppose that $\displaystyle u$ has the mean value property in a domain $\displaystyle \Omega\subset\mathbb{R}^3$ i.e. over any sphere $\displaystyle S(r_0,\delta)$ with interior $\displaystyle B(r_0,\delta)$

$\displaystyle u(r_0)=\frac{1}{4\pi\delta^2}\int_{S(r_0,\delta)}u (r)d\sigma$

For every $\displaystyle \delta>0$ such that $\displaystyle \bar{B}(r_0,\delta)\subset\Omega$. Show that $\displaystyle u$ also has the "volume" mean value property:

$\displaystyle u(r_0)=\frac{1}{\frac{4}{3}\pi\delta^3}\int_{B(r_0 ,\delta)}u(r)dv$

For every ball $\displaystyle B(r_0,\delta)$ such that $\displaystyle \bar{B}(r_0,\delta)\subset\Omega$.

Somehow I think I have to use the fact that $\displaystyle u$ has the mean value property, but I'm not seeing how to do that. I tried comparing the the values in front of the integrands and that was inconclusive (depends on the choice of delta whether one is less than or greater than the other), so I don't know where else to go....