Answer
$\frac{M}{4 \pi R^3(1-2e^{-1})}$
Work Step by Step
We know:
$\rho(R)=\int_0^R \rho_0 e^{\frac{-r}{R}}dr$
We also know that $\rho=\frac{M}{4\pi R^3}$. Thus, we take the integral and simplify to obtain:
$\rho=\frac{M}{4\pi R^3}=\rho_0(1-2e^{-1})$
$\rho_0=\frac{M}{4\pi R^3(1-2e^{-1})}$
