Answer
0.77 A
Work Step by Step
The two resistors $R_{1}$ and $R_{3}$ are in parallel and their combination is
\begin{align*}
R_{\text{13}} = \dfrac{R_{1}R_{3}}{R_{1} + R_{3} } = \dfrac{(45.0 \mathrm{~\Omega})(15.0 \mathrm{~\Omega})}{45.0 \mathrm{~\Omega} + 15.0 \mathrm{~\Omega} } = 11.25 \mathrm{~\Omega}
\end{align*}
This equivalent resistance is in series with $R_2$ and the internal resistance of the battery $r$, so we use Ohm's law to get the current that measured by the ammeter as
\begin{align*}
I = \dfrac{\varepsilon}{R_{13} + R_{2} + r} = \dfrac{25.0 \,\text{V}}{11.25 \mathrm{~\Omega} + 18.0 \mathrm{~\Omega}+ 3.26 \mathrm{~\Omega}} =\boxed{ 0.77 \,\text{A}}
\end{align*}
