Answer
$\sigma = 2.60 \times 10^{-7} \mathrm{C} / \mathrm{m}^{3}$
Work Step by Step
To get the charge density we should get the charge $q$ first by
\begin{align}
q&=E\left(4 \pi r^{2}\right) \epsilon_{0}\\
& =(1750 \mathrm{N} / \mathrm{C})(4 \pi)(0.5 \mathrm{m})^{2}\left(8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}\right)\\
& =4.866 \times 10^{-8} \mathrm{C}
\end{align}
The charge density $\sigma$ equals the charge per unit volume
\begin{align}
\rho&=\frac{q}{V}\\
& =\frac{q}{(4 / 3) \pi R^{3}}\\
& =\frac{4.866 \times 10^{-8} \mathrm{C}}{(4 / 3) \pi(0.355 \mathrm{m})^{3}}\\
& =2.60 \times 10^{-7} \mathrm{C} / \mathrm{m}^{3}
\end{align}
