#### Answer

The sphere's inner diameter is $8.0~cm$

#### Work Step by Step

We can find the volume of aluminum used to make the sphere as:
$V = \frac{M}{\rho}$
$V = \frac{0.690~kg}{2.70\times 10^3~kg/m^3}$
$V = 2.56\times 10^{-4}~m^3$
This volume is equal to the difference between the volume of the outer sphere $V_o$ and the volume of the inner sphere $V_i$. We can find the radius $R_i$ of the inner sphere.
$V_o-V_i = V$
$V_i = V_o-V$
$\frac{4}{3}\pi~R_i^3 = \frac{4}{3}\pi~R_o^3-V$
$R_i^3 = R_o^3-\frac{3V}{4\pi}$
$R_i = (R_o^3-\frac{3V}{4\pi})^{1/3}$
$R_i = [(0.0500~m)^3-\frac{(3)(2.56\times 10^{-4}~m^3)}{4\pi}]^{1/3}$
$R_i = 0.040~m = 4.0~cm$
The sphere's inner diameter is $2R_i$ which is $8.0~cm$.