When the capacitors are connected in series, the resultant capacitance of combination
$$
\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}
$$
$$
\begin{array}{l}
=\frac{1}{1}+\frac{1}{2}+\frac{1}{5}=\frac{17}{10} \\
C=\frac{10}{17} \mu \mathrm{F}
\end{array}
$$
The charge will be same on all the capacitors in series
$$
Q=C V=\frac{10}{17} \times 10=\frac{100}{17}
$$
The potential difference across $2.0 \mu \mathrm{F}$ capacitor
$$
V^{*}=\frac{Q}{C}=\frac{100 / 17}{2}=\frac{50}{17} \text { volt}
$$