Uncertainty in the position of an electron (mass $$= 9.1 \times 10^{-31}\,kg$$) moving with a velocity $$300\,ms^{-1}$$, accurate upto $$0.001\%$$, will be

Heisenberg uncertainty principle: $$\Delta x\,\Delta p \ge \frac{h}{4\pi}$$.

Given velocity $$v = 300\,\text{m s}^{-1}$$ and accuracy $$0.001\%$$.
Percentage $$0.001\% = 0.001/100 = 1 \times 10^{-5}$$.

Uncertainty in velocity:
$$\Delta v = v \times 10^{-5} = 300 \times 10^{-5}\,\text{m s}^{-1} = 3.0 \times 10^{-3}\,\text{m s}^{-1}.$$

Electron mass: $$m = 9.1 \times 10^{-31}\,\text{kg}$$.
Uncertainty in momentum:
$$\Delta p = m\,\Delta v = (9.1 \times 10^{-31})(3.0 \times 10^{-3})$$
$$\;\;\; = 2.73 \times 10^{-33}\,\text{kg m s}^{-1}.$$

Hence uncertainty in position:
$$\Delta x = \frac{h}{4\pi\,\Delta p} = \frac{6.63 \times 10^{-34}}{4\pi \times 2.73 \times 10^{-33}}$$
$$= \frac{6.63 \times 10^{-34}}{12.56 \times 2.73 \times 10^{-33}}$$
$$= \frac{6.63}{34.3} \times 10^{-34+33} \approx 1.93 \times 10^{-2}\,\text{m}.$$

Therefore the uncertainty in position is $$\boxed{1.92 \times 10^{-2}\,\text{m}}$$.

Option C which is: $$1.92 \times 10^{-2}\,m$$.