Answer
$q = 7.78\times 10^{-15}~C$
Work Step by Step
We can use an integral to find the total charge:
$q = \int_{0}^{R}~\rho~4\pi~r^2~dr$
$q = \int_{0}^{R}~\frac{(14.1~pC/m^3)~r}{R}~4\pi~r^2~dr$
$q = \int_{0}^{R}~\frac{(14.1~pC/m^3)~4\pi~r^3}{R}~dr$
$q = \frac{(14.1~pC/m^3)~4\pi~r^4}{4R} ~\Big \vert_{0}^{R}$
$q = \frac{(14.1~pC/m^3)~4\pi~R^4}{4R} -0$
$q = \frac{(14.1~pC/m^3)~4\pi~R^3}{4}$
$q = \frac{(14.1\times 10^{-12}~C/m^3)~(4\pi)~(0.0560~m)^3}{4}$
$q = 7.78\times 10^{-15}~C$
