1. Let each capacitor have capacitance $C = 4\,\mu F$

2. Check each option.

Option A: all in series

For three equal capacitors in series, $\frac{1}{C_{\text{eq}}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$ So, $C_{\text{eq}} = \frac{4}{3}\,\mu F$ This is not $6\,\mu F$.

Option B: two in series and one in parallel

First, combine two capacitors of $4\,\mu F$ in series: $C_s = \frac{4\times 4}{4+4} = \frac{16}{8} = 2\,\mu F$ Now this combination is in parallel with the third $4\,\mu F$ capacitor: $C_{\text{eq}} = 2 + 4 = 6\,\mu F$ This matches the required value.

Option C: all in parallel

For three equal capacitors in parallel, $C_{\text{eq}} = 4+4+4 = 12\,\mu F$ This is not $6\,\mu F$.

Option D: two in parallel and one in series

First, combine two capacitors in parallel: $C_p = 4+4 = 8\,\mu F$ Now this is in series with the third $4\,\mu F$ capacitor: $C_{\text{eq}} = \frac{8\times 4}{8+4} = \frac{32}{12} = \frac{8}{3}\,\mu F$ This is not $6\,\mu F$.

3. Therefore, the only correct arrangement is: $\boxed{\text{B: two in series and one in parallel}}$

4. Comparison with stored correct answer:

• Stored correct answer: B
• Derived answer: B

So, the derived answer agrees with the stored correct answer.