All the capacitors are in parallel combination and therefore potential drop across each capacitor would be the same.

Charge on the capacitor can be written as, $Q=CV$.

The charge on $C_1$ is $Q_1=2\times 10=20 \mathrm{\mu C}$

The charge on $C_2$ is $Q_2=x\times 10=10x \mathrm{\mu C}$

The charge on $C_3$ is $Q_3=3\times 10=30 \mathrm{\mu C}$

By the given data.

$C_1+C_2+C_3=100 \mathrm{\mu C}$

$\Rightarrow 10\left(2+x+3\right)=100$

$\Rightarrow x=5$