Answer
$6.1\times10^{-5}\,C$
Work Step by Step
Recall: $E=\frac{qV}{2}$ or $V=2E/q$
Therefore, for capacitor A, we have
$V=\frac{2\times5.0\times10^{-5}\,J}{11\times10^{-6}\,C}=9.1\,V$
The same voltage is applied between the plates of capacitor B, i.e., $V_{B}=9.1\,V$.
Capacitance of capacitor B, $C_{B}=6.7\,\mu F=6.7\times10^{-6}\,F$
Then, $q_{B}=C_{B}V_{B}=6.7\times10^{-6}\,F\times9.1\,V=6.1\times10^{-5}\,C$
