Answer
The direction of the current through resistance 1 is down.
Work Step by Step
Let $i_1$ be the current through resistance 1 and let's assume the direction of the current is down.
Let $i_2$ be the current through resistance 2 and let's assume the direction of the current is to the left.
Let $i_3$ be the current through battery 2 and let's assume the direction of the current is up.
Using the junction rule:
$i_3 = i_1+i_2$
Using the loop rule counterclockwise around the right loop:
$12.0-200~i_1 = 0$
$200~i_1 = 12.0$
$i_1 = \frac{12.0}{200}$
$i_1 = 60.0~mA$
Using the loop rule counterclockwise around the left loop:
$6.00+12.0-100~i_2 = 0$
$100~i_2 = 18.0$
$i_2 = 180~mA$
The current through resistance 1 is $~~60.0~mA$
We assumed that the direction of the current through resistance 1 is down, and we found the current to be $~~60.0~mA$
Since the value we found for $i_1$ is positive, the direction of the current through resistance 1 is down.
