Answer
$\rho=-1.3\times 10^{-8}\frac{C}{m^3}$
Work Step by Step
We can find the total area surface bounding the bathroom as
$A=2(2.5\times 3.0)+2(3.0\times 2.0)+2(2.0\times 2.5)=37m^2$
We know that:
$\phi=E.A$
We plug in the known values to obtain:
$\phi=(600)(37)=22\times 10^3\frac{N.m^2}{C}$
Now, according to Gauss's law
$q=\epsilon_{\circ}\phi$
We plug in the known values to obtain:
$q=(8.85\times 10^{-12})(22\times 10^3)=2.0\times 10^{-7}C$
As the charges are negative, so:
$q=-2.0\times 10^{-7}C$
Now,
$\rho=\frac{q}{V}$
$\rho=\frac{-2.0\times 10^{-7}}{15}=-1.3\times 10^{-8}\frac{C}{m^3}$
