b
Here the three plates will make ( $3-1$ ) $= 2$ capacitors of same capacitance $C$ and these two capacitors are connected in parallel.
here, $C=\frac{A \in_{0}}{d}=\frac{0.1 \times 8.854 \times 10^{-12}}{0.885 \times 10^{-3}}=1 n F$
The equivalent capacitance of the system is $C_{e q}=C+C=2 n F$
Energy stored in this system is $U=\frac{1}{2} C_{e q} V^{2}=\frac{1}{2} \times 2 \times 10^{-9} \times 10^{2}=10^{-7} J=10^{-1} \mu J$