Three capacitances, each of 3 $$\mu$$F, are provided. These cannot be combined to provide the resultant capacitance of:

Each capacitor has capacitance $$C = 3\,\mu\text{F}$$.
To obtain an equivalent capacitance we must connect all three capacitors, using any series-parallel arrangement. With three identical capacitors there are only four essentially different ways to combine them.

All three in series.
$$C_{\text{eq}} = \frac{C}{3} = \frac{3}{3} = 1\,\mu\text{F}$$

All three in parallel.
$$C_{\text{eq}} = 3C = 3 \times 3 = 9\,\mu\text{F}$$

Two in series, that pair in parallel with the third.
Series pair: $$C_s = \frac{C}{2} = \frac{3}{2} = 1.5\,\mu\text{F}$$
Putting this in parallel with the remaining $$3\,\mu\text{F}$$ capacitor:
$$C_{\text{eq}} = C_s + C = 1.5 + 3 = 4.5\,\mu\text{F}$$

Two in parallel, that pair in series with the third.
Parallel pair: $$C_p = 2C = 2 \times 3 = 6\,\mu\text{F}$$
Putting this in series with the remaining $$3\,\mu\text{F}$$ capacitor:
$$\frac{1}{C_{\text{eq}}} = \frac{1}{C_p} + \frac{1}{C} = \frac{1}{6} + \frac{1}{3} = \frac{1}{6} + \frac{2}{6} = \frac{3}{6}$$
so $$C_{\text{eq}} = \frac{6}{3} = 2\,\mu\text{F}$$

Thus the only equivalent capacitances obtainable with all three $$3\,\mu\text{F}$$ capacitors are

$$1\,\mu\text{F},\;2\,\mu\text{F},\;4.5\,\mu\text{F},\;9\,\mu\text{F}$$

From the given options the value that cannot be produced is
Option B: $$6\,\mu\text{F}$$.