Since the question raised a discussion in the comments, I guess it won't hurt putting up a brief summarizing answer.
Notation % m/v, as you might've guessed, refers to the ratio between the mass and the volume, or, more precisely, mass concentration $ρ_i$ (sometimes denoted as $γ_i$ [1, p. 48] to distinguish from density, see How to distinguish mass concentration and density?):
$$ρ_i = \frac{m_i}{V}$$
where $m_i$ is the mass of solute and $V$ is the volume of solution, suggesting the SI unit $\pu{kg m-3}.$
Where does percent in % m/v come from?
Quoting Wikipedia:
In biology, the "%" symbol is sometimes incorrectly used to denote mass concentration, also called "mass/volume percentage." A solution with 1 g of solute dissolved in a final volume of 100 mL of solution would be labeled as "1%" or "1% m/v" (mass/volume). The notation is mathematically flawed because the unit "%" can only be used for dimensionless quantities.
So, in order to account for this percentage “unit”, one should normalize the mass concentration to the mass of solute in grams in 100 mL of solution.
This can be done in one's head in most cases, but you must watch out for the units.
In this particular example, only option b satisfy the definition of the % m/v quantity we are expressing since only b lists volume of KI solution.
Let's check the math.
Since $\pu{575.0 mg} = \pu{0.5750 g},$ there is $\pu{0.5750 g}$ $\ce{KI}$ per $\pu{10.00 mL}$ of $\ce{KI}$ solution, and normalizing to $\pu{100 mL}$ we indeed obtain the aforementioned concentration:
$$ρ_\ce{KI} = \frac{\pu{0.5750 g}}{\pu{10.00 mL}} = \frac{\pu{5.750 g}}{\pu{100.0 mL}} = 5.750\%~(\text{m/v})$$
The exercise really just requires to pay more attention and follow the analytical chemistry definitions.
References
- IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
