1. Condition for balance of Wheatstone bridge

For a balanced Wheatstone bridge,

$\frac{R_1}{R_2} = \frac{R_3}{R_4}$

or equivalently,

$R_3 = R_4\cdot \frac{R_1}{R_2}$

2. Decode the value of $R_1$

Given colour code for $R_1$: Orange, Red, Brown.

Using resistor colour code:

• Orange $= 3$
• Red $= 2$
• Brown multiplier $= 10$

So,

$R_1 = 32 \times 10 = 320\ \Omega$

3. Substitute the given values

Given:

• $R_2 = 80\ \Omega$
• $R_4 = 40\ \Omega$

Thus,

$R_3 = 40 \cdot \frac{320}{80}$

$R_3 = 40 \cdot 4 = 160\ \Omega$

4. Find the colour code for $160\ \Omega$

We write:

$160 = 16 \times 10$

Now,

• 1 $\to$ Brown
• 6 $\to$ Blue
• multiplier $10$ $\to$ Brown

So the colour code is:

Brown, Blue, Brown

5. Check options

• A: Brown, Blue, Brown $\rightarrow 160\ \Omega$ ✅
• B: Grey, Black, Brown $\rightarrow 80 \times 10 = 800\ \Omega$
• C: Red, Green, Brown $\rightarrow 25 \times 10 = 250\ \Omega$
• D: Brown, Blue, Black $\rightarrow 16 \times 1 = 16\ \Omega$

Therefore, the correct option is A.