The de Broglie wavelength of an oxygen molecule at $$27^{\circ}C\text{ is }x\times10^{-12}m$$. The value of x is (take Planck's constant $$=6.63\times10^{-34}J.s$$, Boltzmann constant $$=1.38\times10^{-23}J/K$$, mass of oxygen molecule $$=5.31\times10^{-26}kg$$)

We need to find the de Broglie wavelength of an O$$_2$$ molecule at 27°C. The de Broglie wavelength of a gas molecule at temperature T is: $$\lambda = \frac{h}{\sqrt{3mk_BT}}$$ where $$h$$ is Planck's constant, $$m$$ is the mass of one molecule, $$k_B$$ is Boltzmann's constant, and $$T$$ is the absolute temperature.

$$h = 6.63 \times 10^{-34} \, \text{J·s}$$
$$m = 32 \times 1.66 \times 10^{-27} \, \text{kg} = 53.12 \times 10^{-27} \, \text{kg}$$ (O$$_2$$ has molar mass 32 g/mol)
$$k_B = 1.38 \times 10^{-23} \, \text{J/K}$$
$$T = 27°C = 300 \, \text{K}$$

$$3mk_BT = 3 \times 53.12 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300$$
$$= 3 \times 53.12 \times 1.38 \times 300 \times 10^{-50}$$
$$= 3 \times 21992.64 \times 10^{-50} = 65977.9 \times 10^{-50}$$
$$= 6.598 \times 10^{-46}$$

$$\sqrt{3mk_BT} = \sqrt{6.598 \times 10^{-46}} = 2.569 \times 10^{-23}$$

$$\lambda = \frac{6.63 \times 10^{-34}}{2.569 \times 10^{-23}} = 2.581 \times 10^{-11} \, \text{m} = 25.81 \times 10^{-12} \, \text{m} \approx 26 \times 10^{-12} \, \text{m}$$

So $$x = 26$$.

The correct answer is Option (4): 26.