Answer
$q=11nC$
Work Step by Step
The free charge, $q$, before the slab is inserted can be calculated as
$q=C_{\circ}V$
We plug in the known values to obtain:
$q=89\times10^{-12}(120)$
$q=10.680\times10^{3}\times10^{-12}=10.680\times10^{-9}$
$q=11\times10^{-9}$
$q=11nC$
