Answer
$E=4.0\times 10^6\frac{N}{C}$
Work Step by Step
The direction of electric field at $P_1$ is away from $q_1$ and we can find its magnitude as:
$E=\frac{1 }{4\pi \epsilon_{\circ}}\frac{q}{r^2}$
We plug in the known values to obain:
$E=8.99\times 10^9\times \frac{1.0\times 10^{-7}}{(0.015)^2}=4.0\times 10^6\frac{N}{C}$
