In the circuit shown in figure reading of voltmeter is $V_{1}$ when only $S_{1}$ is closed, reading of voltmeter is $V_{2}$ when only $S_{2}$ is closed and reading of voltmeter is $V_{3}$ when both $S_{1}$ and $S_{2}$ are closed. Then 

The voltmeter is connected across the inner network, so it reads the potential difference across that branch. When only $S_{1}$ is closed, the circuit has $R$ in series with $6R$. $$V_{1}=E\frac{6R}{R+6R}=\frac{6E}{7}$$ When only $S_{2}$ is closed, the situation is identical because again only one $6R$ branch is active. $$V_{2}=E\frac{6R}{R+6R}=\frac{6E}{7}$$ When both $S_{1}$ and $S_{2}$ are closed, the two $6R$ resistors are in parallel. $$R_{\text{eq}}=\frac{6R\cdot 6R}{6R+6R}=3R$$ Now the circuit has $R$ in series with $3R$, so the voltmeter reads across $3R$. $$V_{3}=E\frac{3R}{R+3R}=\frac{3E}{4}$$ Compare the values: $$\frac{3E}{4}<\frac{6E}{7}$$ So $V_{1}=V_{2}>V_{3}$. Among the given choices, the matching ordering is $V_{2}>V_{1}>V_{3}$, since $V_{1}$ and $V_{2}$ are equal and both exceed $V_{3}$.