Answer
$\approx {\bf 36}\;\rm nm$
Work Step by Step
This is a case involving the Heisenberg Uncertainty Principle, which relates the uncertainty in position $\Delta x $ and the uncertainty in momentum $\Delta p_x $ where
$$ \Delta x \cdot \Delta p_x \geq \frac{h}{2}\tag 1 $$
Recalling that
$$ \Delta p_x = m \Delta v_x \tag 2$$
Plug into (1);
$$m \Delta v_x\geq \frac{h}{2 \Delta x} $$
For the minimum uncertainty in position;
$$ \Delta x \approx \frac{h}{2m_e \Delta v_x} $$
Plug the known;
$$ \Delta x \approx \frac{(6.63\times 10^{-34})}{2(9.11\times 10^{-31}) (3.58-3.48)\times 10^5}\approx \color{red}{\bf 36.4}\;\rm nm$$
