Answer
See the detailed answer below.
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 range of speeds
$$ \Delta v_x \geq \frac{h}{2m_p \Delta x} $$
Plug the known;
$$ \Delta v_x \geq \frac{(6.63\times 10^{-34})}{2(1.67\times 10^{-27}) (4\times 10^{-15})}= {\bf 5\times 10^7}\;\rm m/s$$
And since the average velocity is zero, the proton's velocity range is then
$$ -2.5\times 10^{7}\;{\rm m/s}\leq 2.5\times 10^{-7}\leq \;{\rm m/s }$$
Hence the smallest range of proton's speed is from $0$ to $2.5\times 10^7$ m/s.
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Note: the nucleus diameter is given as 4.0 m in your textbook which is a copy mistake, the nucleus diameter is measured in femto meters as we substituted above.
