Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
When the swithch is open, there is no current passes through the right loop.
The current is given by applying Ohm's law
$$\varepsilon=IR_{eq}$$
where $R_{eq}=R+r$, so
$$I_A=\dfrac{\varepsilon}{R+r}$$
where $I$ is the current that comes out from the battery and, in this case, it is equal to the current that goes through $R$.
Plug the known;
$$I_A=\dfrac{1.5}{6+0.5}=\frac{3}{13}\;\rm A=\color{red}{\bf 0.2308}\;\rm A$$
$$\color{blue}{\bf [b]}$$
When the swithch is closed,
The battery current is given by applying Ohm's law
$$\varepsilon=IR_{eq}$$
where $R_{eq}=\left[R_A^{-1}+R_B^{-1}\right]^{-1}+r$, so
$$I =\dfrac{\varepsilon}{\left[R^{-1}+R^{-1}\right]^{-1}+rr}$$
Plug the known;
$$I =\dfrac{1.5}{\left[6^{-1}+6^{-1}\right]^{-1}+0.5}=\bf \frac{3}{7}\;\rm A$$
where
$$I=I_A+I_B$$
where the two resistors are in parallel, so
$$\Delta V_A=\Delta V_B$$
$$I_AR_A=I_BR_B$$
Thus,
$$I_A=I_B$$
And therefore,
$$I=2I_A $$
$$I_A=\dfrac{I}{2}=\dfrac{3}{7(2)}$$
$$I_A=\color{red}{\bf 0.2143}\;\rm A$$
$$\color{blue}{\bf [c]}$$
The percentage of change is given by
$$=\dfrac{I_{A2}-I_{A1}}{I_{A1}}\times 100\% =\dfrac{0.2143-0.2308}{0.2308}\times 100\%=-\color{red}{\bf 7.15}\%$$
The current decreases by 7%.
