Answer
(a) greater than
(b) $-4.50\mu C$
Work Step by Step
(a) We know that
$q=\frac{mgsin\theta}{Ecos\theta}$
$\implies q=\frac{mg}{E} tan\theta$
Thus, the magnitude of new $q$ is greater than the previous values of $q$ because the angle $\theta$ is greater than the previous case.
(b) We know that
$q=\frac{mg}{E}tan\theta$
We plug in the known values to obtain:
$q=\frac{(0.025Kg)(9.8m/s^2)}{1.46\times 10^4N/C}tan15^{\circ}$
$\implies q=-4.5\times 10^6C=-4.50\mu C$