Answer
$C_{\circ}=89pF$
Work Step by Step
The capacitance before the slab is inserted is
$C_{\circ}=\epsilon_{\circ}\frac{A}{d}$
We plug in the known values to obtain:
$C_{\circ}=8.85\times10^{-12}\frac{0.12}{1.2\times10^{-2}}$
$C_{\circ}=8.85\times10^{-11}$
We round to simplify:
$C_{\circ}=8.9\times10^{-11}$
$C_{\circ}=89\times10^{-12}$
$C_{\circ}=89pF$
