I think this is just a simple application of the Heisenberg uncertainty principle.
There are multiple ways to express the Heisenberg uncertainty principle. One way is in terms of an approximate formula involving uncertainties in terms of "deltas." Another form is a very precise inequality that involves standard deviations, and its use generally requires accurate information about the specific wavefunction shape.
I'm guessing this problem involves the much easier approximation with the "delta" uncertainties. If you use that one, the answer is one of the listed choices.
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Edit: But again, there are different ways to express the Heisenberg uncertainty principle. In its simplest form,
[tex]\Delta p \Delta x \approx h[/tex]
But then there is a slightly more approximate inequality,
[tex]\Delta p \Delta x \gtrsim h[/tex]
And to make things more confusing, that's sometimes expressed by
[tex]\Delta p \Delta x \gtrsim \frac{h}{2}[/tex]
But the most exact version requires you know quite a bit about the wavefunction. You can calculate the variance of the position and variance of the momentum using standard quantum mechanical operators. For example, assuming the expectation value of position and momentum are both 0, then in 1-demension,
[tex]\sigma_x^2 = \int_{-\infty}^{\infty} \psi^* x^2 \psi \ dx[/tex]
[tex]\sigma_p^2 = \int_{-\infty}^{\infty} \psi^* \left( -\hbar^2 \frac{\partial^2}{\partial x^2} \right) \psi \ dx[/tex]
Then in terms of standard deviations, one can show the ultimate in the uncertainty principle:
[tex]\sigma_x \sigma_p \geq \frac{\hbar}{2}[/tex]
where [itex]\hbar[/itex] = h/(2π)
I'm guessing the version you are supposed to use is one of the first three. Check your textbook/coursework for the preferred version in your course.