The connecting wires and the battery in the circuit shown below have negligible resistance.  The current (amperes) through the $$6Omega$$ resistor is most nearly:

We can solve this problem by first taking a look at Ohm's law, which states:

$V = I*R$

Using this equation, we can write two separate expressions for the voltage across both the  3Ω and 6Ω resistors.

$V_{3\Omega }=\left(I_{3\Omega }\right)\left(3\Omega \right)$ 

$V_{6\Omega }=\left(I_{6\Omega }\right)\left(6\Omega \right)$ 

Since the voltage across parallel resistors is equal:

$V = V_{3\Omega }=V_{6\Omega }$

$\left(I_{3\Omega }\right)\left(3\Omega \right) = \left(I_{6\Omega }\right)\left(6\Omega \right)$ 

$\left(I_{3\Omega }\right) = 2\left(I_{6\Omega }\right)$

The above expression states that the current going through the 3Ω resistor is 2 times greater than that going through the 6Ω. Since all current must flow through either one of the two resistors as it makes its way around the circuit, we know that:

$I_{total = }{I}_{3\Omega } + I_{6\Omega } = 2\left(I_{6\Omega }\right) + I_{6\Omega } = 3\left(I_{6\Omega }\right)$ 

Now we must solve for $I_{total}$, using the equation:

$V_{total }=\left(I_{total}\right)\left(R_{total}\right)$

To get $R_{total}$, we must simplify the 3 resistors into one equivalent resistance. Resistors in parallel can be combined according to the following equation:

$R_{eq}=\frac{\left(R_1\right)\left(R_2\right)}{R_1+R_2}$

The equivalent resistance of the 3Ω and 6Ω resistors is then equal to:

$R_{eq}=\frac{\left(3\right)\left(6\right)}{3+6}=\frac{18}{9}=2\Omega$

Since resistors in series combine, we simply add $R_{eq}$ to the 4Ω resistor to get an $R_{total}$ value of 6Ω. Plugging this back into the equation along with $V_{total }= 6V$ gives us a value of 1A for $I_{total}$:

$6V=\left(I_{total}\right)\left(6\Omega \right)$

$I_{total} = 1A$

Finally, we plug this value in to get our current through our 6Ω resistor:

$I_{total = } 3\left(I_{6\Omega }\right)$

$1A = 3\left(I_{6\Omega }\right)$

$I_{6\Omega } = \frac{1}{3}A$