Answer
$Q=8.2\times10^{-7}C$; positive
Work Step by Step
Electric fields point toward the negative and the charge is repelled by it, so it must be a positive charge.
$\cos(\theta)=\frac{55cm-12cm}{55cm}$
$\theta=38.6^o$
$mg=(0.001kg)(9.81\frac{m}{s^2})=0.00981N$
$\tan(\theta)=\frac{F}{mg}$
$F=mg\tan(\theta)=(0.00981N)\tan(38.6^o)=0.00782N$
$Q=\frac{F}{E}=\frac{0.00782N}{9500N/C}=8.2\times10^{-7}C$; positive
