#### Answer

(a) $E = 36~N/C$
(b) $E = 18~N/C$

#### Work Step by Step

(a) We can find the electric field strength at the point $(10~cm, 0)$ which is along the axis of the dipole.
$E = \frac{q~d}{2\pi~\epsilon_0~r^3}$
$E = \frac{(1.0\times 10^{-9}~C)(0.0020~m)}{(2\pi)(8.85\times 10^{-12}~C^2/N~m^2)(0.10~m)^3}$
$E = 36~N/C$
(b) We can find the electric field strength at the point $(0,10 ~cm)$ which is along the y-axis.
$E = \frac{q~d}{4\pi~\epsilon_0~r^3}$
$E = \frac{(1.0\times 10^{-9}~C)(0.0020~m)}{(4\pi)(8.85\times 10^{-12}~C^2/N~m^2)(0.10~m)^3}$
$E = 18~N/C$