Answer
$a) \space q = +10.9\mu C $
$b) \space$ $F = 0.600N$ directed downwards
Work Step by Step
$a)$ Since the charge is below the $-550C$ charge and is attracted towards it up. The sign of charge should be opposite to $-550 C$ charge and hence should be positive..
$F={kq_1q_2\over r^2}$
$q_1 = -0.550
\mu C, q_2=q, F=0.600N, r = 0.300m$
$q=\big|{Fr^2\over kq_1}\big| =\big|{0.6 (0.3)^2\over 9\times 10^9\times 0.550 \times 10^{-6}}\big| $
$q = 10.9\mu C$
$b)$ Since the distance is still the same, the magnitude of the Force would be the same and directed downwards i.e $F = 0.600N$ directed downwards
