#### Answer

(a) $q = 2.0~nC$
(b) $E = 180~N/C$

#### Work Step by Step

(a) The point $(0~cm, 10~cm)$ is along the axis of the dipole. We can find the charge $q$.
$E = \frac{q~d}{2\pi~\epsilon_0~r^3}$
$q = \frac{2\pi~\epsilon_0~r^3~E}{d}$
$q = \frac{(2\pi)(8.85\times 10^{-12}~C^2/N~m^2)(0.10~m)^3(360~N/C)}{0.010~m}$
$q = 2.0~nC$
(b) We can find the electric field strength at the point $(10 ~cm, 0~cm)$ which is along the x-axis.
$E = \frac{q~d}{4\pi~\epsilon_0~r^3}$
$E = \frac{(2.0\times 10^{-9}~C)(0.0010~m)}{(4\pi)(8.85\times 10^{-12}~C^2/N~m^2)(0.10~m)^3}$
$E = 180~N/C$