Step 1: Plan:
We find the current in the $3\,\Omega$ resistor with Thevenin's theorem.
We remove the $3\,\Omega$ resistor, find the open-circuit voltage between T and B, then find the resistance seen between T and B.
Step 2: Open-circuit voltage between T and B:
Put the left corner at $6\,\text{V}$ and the right corner at $0\,\text{V}$. Each side is then a simple potential divider.
\[
V_T=6\times\frac{2}{20+2}=\frac{6}{11}\,\text{V},\qquad V_B=6\times\frac{1}{12+1}=\frac{6}{13}\,\text{V}
\]
\[
V_{th}=V_T-V_B=\frac{6}{11}-\frac{6}{13}=\frac{12}{143}\,\text{V}\approx 0.084\,\text{V}
\]
The bridge is not balanced, since $\frac{20}{2}\neq\frac{12}{1}$.
Step 3: Thevenin resistance:
Replace the 6 V cell by a wire. The left and right corners are now at the same point.
Seen from T, the $20\,\Omega$ and $2\,\Omega$ arms are in parallel. Seen from B, the $12\,\Omega$ and $1\,\Omega$ arms are in parallel.
\[
R_T=\frac{20\times2}{22}=\frac{20}{11}\,\Omega,\qquad R_B=\frac{12\times1}{13}=\frac{12}{13}\,\Omega
\]
\[
R_{th}=R_T+R_B=\frac{20}{11}+\frac{12}{13}=\frac{392}{143}\,\Omega\approx 2.74\,\Omega
\]
Step 4: Current through the load:
The $3\,\Omega$ resistor is the load across the Thevenin source.
\[
I=\frac{V_{th}}{R_{th}+3}=\frac{12/143}{(392+429)/143}=\frac{12}{821}\,\text{A}
\]
\[
I\approx 1.46\times10^{-2}\,\text{A}
\]
Step 5: Meaning of the result:
The current is small because the bridge is nearly balanced, but it is not zero.
The direction is from T to B if the left corner is the positive terminal.
Final Answer:
\[
\boxed{I_{3\,\Omega}=\frac{12}{821}\,\text{A}\approx 1.46\times10^{-2}\,\text{A}}
\]
