Calculate the current across the 8 ohm resistor

Let the bottom wire be the ground ($0 \, V$).
Let $V_A$ be the voltage at the node after the $2 \, \Omega$ resistor ($R_6$) and before the $8 \, \Omega$ ($R_5$) and $12 \, \Omega$ ($R_4$) resistors.
Let $V_B$ be the voltage at the node after the $8 \, \Omega$ resistor ($R_5$) and before the $6 \, \Omega$ resistor ($R_1$).
Let $V_C$ be the voltage at the node after the $12 \, \Omega$ resistor ($R_4$) and before the $4 \, \Omega$ resistor ($R_3$).
Let $V_D$ be the voltage at the node after the $6 \, \Omega$ resistor ($R_1$) and before the $3 \, \Omega$ resistor ($R_2$).
The voltage source is $28 \, V$.

KCL at Node A:

$$\frac{28 \, V - V_A}{2 \, \Omega} = \frac{V_A - V_B}{8 \, \Omega} + \frac{V_A - V_C}{12 \, \Omega}$$

Multiply by 24 (LCM of 2, 8, 12):

$$12(28 - V_A) = 3(V_A - V_B) + 2(V_A - V_C)$$ $$336 - 12V_A = 3V_A - 3V_B + 2V_A - 2V_C$$ $$336 = 17V_A - 3V_B - 2V_C \quad (\text{Equation 1})$$

KCL at Node B:

$$\frac{V_A - V_B}{8 \, \Omega} = \frac{V_B - V_D}{6 \, \Omega}$$

Multiply by 24:

$$3(V_A - V_B) = 4(V_B - V_D)$$ $$3V_A - 3V_B = 4V_B - 4V_D$$ $$3V_A = 7V_B - 4V_D \quad (\text{Equation 2})$$

KCL at Node C:

$$\frac{V_A - V_C}{12 \, \Omega} = \frac{V_C - 0 \, V}{4 \, \Omega}$$

Multiply by 12:

$$V_A - V_C = 3V_C$$ $$V_A = 4V_C \quad (\text{Equation 3})$$

KCL at Node D:

$$\frac{V_B - V_D}{6 \, \Omega} = \frac{V_D - 0 \, V}{3 \, \Omega}$$

Multiply by 6:

$$V_B - V_D = 2V_D$$ $$V_B = 3V_D \quad (\text{Equation 4})$$

From Equation 3, $V_C = \frac{1}{4}V_A$.
From Equation 4, $V_D = \frac{1}{3}V_B$.

Substitute $V_D$ into Equation 2:

$$3V_A = 7V_B - 4\left(\frac{1}{3}V_B\right)$$ $$3V_A = 7V_B - \frac{4}{3}V_B$$ $$3V_A = \frac{21V_B - 4V_B}{3}$$ $$3V_A = \frac{17}{3}V_B$$ $$9V_A = 17V_B \implies V_B = \frac{9}{17}V_A \quad (\text{Equation 5})$$

Now substitute $V_C$ and $V_B$ (in terms of $V_A$) into Equation 1:

$$336 = 17V_A - 3\left(\frac{9}{17}V_A\right) - 2\left(\frac{1}{4}V_A\right)$$ $$336 = 17V_A - \frac{27}{17}V_A - \frac{1}{2}V_A$$

Find a common denominator (34) for the coefficients of $V_A$:

$$336 = V_A \left(\frac{17 \times 34}{34} - \frac{27 \times 2}{34} - \frac{1 \times 17}{34}\right)$$ $$336 = V_A \left(\frac{578 - 54 - 17}{34}\right)$$ $$336 = V_A \left(\frac{507}{34}\right)$$ $$V_A = \frac{336 \times 34}{507} = \frac{11424}{507} \, V$$

The current across the $8 \, \Omega$ resistor ($R_5$) is $I_{R5} = \frac{V_A - V_B}{8 \, \Omega}$.
First, find the voltage difference $V_A - V_B$:

$$V_A - V_B = V_A - \frac{9}{17}V_A = \left(1 - \frac{9}{17}\right)V_A = \frac{17 - 9}{17}V_A = \frac{8}{17}V_A$$

Now substitute $V_A$:

$$V_A - V_B = \frac{8}{17} \times \frac{11424}{507} = \frac{8 \times 11424}{17 \times 507} = \frac{91392}{8619} \, V$$

Finally, calculate the current $I_{R5}$:

$$I_{R5} = \frac{V_A - V_B}{8} = \frac{\frac{91392}{8619}}{8} = \frac{91392}{8 \times 8619} = \frac{11424}{8619} \, A$$

Simplify the fraction by dividing by common factors. Both numerator and denominator are divisible by 3:

$$\frac{11424 \div 3}{8619 \div 3} = \frac{3808}{2873}$$

Both are divisible by 17:

$$\frac{3808 \div 17}{2873 \div 17} = \frac{224}{169}$$

The fraction cannot be simplified further as $169 = 13^2$ and 224 is not divisible by 13.

The current across the $8 \, \Omega$ resistor is $\boxed{\frac{224}{169} \, \text{A}}$.