How to apply Faraday's law to calculate the charge of metal ions?
When you're confused by a problem like this it is definitely time to use dimensional analysis.

For the amount of electrons $\text{mol}_e$

\begin{align}
\text{The ampere} &amp;= \text{Amp} = 1.00 \frac{\text{coulomb}}{\text{second}} \approx 6.242\times10^{18}~\frac{\text{electrons}}{\text{second}}\\
\text{Farady} &amp;= 26.801 \frac{\text{Amp}\times\text{hours}}{\text{mol}_e}\\
30~\text{minutes} &amp;= \frac{30 \text{minutes}}{60~\text{minutes/hour}} = 0.500~\text{hours}\\
\text{mol}_e &amp;= \frac{0.50~\text{Amp} \times 0.500~\text{hours}}{26.801~\text{Amp}\times\text{hours}\text{/mol}_e} = 9.328\times10^{-3}~\text{mol}_e
\end{align}

For the amount of substance of metal $\text{m}_m$

$$\text{mol}_m = \frac{0.700~\text{g}}{150~\text{g/mol}_m} = 4.667\times10^{-3}~\text{mol}_m$$

Charge per atom = $z$

$$z = \frac{\text{mol}_e}{\text{mol}_m}= \frac{9.328\times10^{-3}}{4.667\times10^{-3}} = 1.998 = 2.0 \frac~{\text{electrons}}{\text{atom}}$$