I know this question was over 2 years ago, but I've just joined and wanted to see if I could improve on the responses already given and practice the formatting of text while also perhaps helping someone else who might come along this type of problem.
Acetic acid is a weak acid, as others have mentioned. Acetic acid does not fully ionize in water. $\mathrm{p}K_\mathrm{a}$ becomes necessary here.
The $\mathrm{p}K_\mathrm{a}$ of acetic acid is $4.76$. You should have been permitted to locate this value in a table or otherwise been provided it on the test form, unless you were required to memorize acid dissociation constants, in which case I apologize for what must have been a course destined to forever char (oxidation!) your memory and appreciation for the magical world of chemistry.
From the definition of $\mathrm{p}K_\mathrm{a}$ we see that $$\mathrm{p}K_\mathrm{a} = -\lg K_\mathrm{a}$$ which implies that the $K_\mathrm{a}$ of acetic acid is $10^{-4.76}$, which is $1.74 \times 10^{-5}$.
Now, consider that,
$$K_\mathrm{a} = \frac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$
We are asked for $\mathrm{pH}$ in the problem, and we know that:
$$\mathrm{pH} = -\lg [\ce{H^+}]$$
So let's work on isolating $[\ce{H^+}]$ so that we can get the $\mathrm{pH}$ for our answer.
Let's consider it a given that our acetic acid is in pure water which means, when it reaches equilibrium, it does so to equimolar portions of protons and acetate anions, viz.,
$$\ce{HA<<=>H^+ + A^-}$$ where the $K_\mathrm{a}$ of this equilibrium we determined above to be $1.74 \times 10^{-5}$.
As we mentioned above, $$K_\mathrm{a} = \frac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$
We talked about the very reasonable assumption that $[\ce{H^+}]$ and $[\ce{A^-}]$ are in equimolar proportions on dissociation of $[\ce{HA}]$. If we let $$x = [\ce{H^+}] = [\ce{A^-}]$$ then it follows that:
$$K_\mathrm{a} = \frac{x^2}{[\ce{HA}]}$$
Since the $K_\mathrm{a}$ of acetic acid is $1.74 \times 10^{-5}$, what are we left with? What do we know? That $5~\%$ is coming into play now I think. We know that per cent is "per 100," but "per 100" of what? In chemistry, in the absence of specific details otherwise, my assumption is molarity (moles/litre). Molality is used at times (mass fraction) as may be volume and weight percentages, mole fractions, etc. Given just the "$5~\%$" statement in the problem, I would take that as molarity of the acid ($[\ce{HA}]$).
Molarity is concentration! So the equations above, particularly the one that equates $K_\mathrm{a}$ to concentrations of products/reactants should work!
If $1.74 \times 10^{-5} = K_\mathrm{a}$ and $0.05 = [\ce{HA}]$, then it follows that
$$x = [\ce{H^+}] = \sqrt{1.74 \times 10^{-5} \times 0.05}$$
On my calculator, I get $x = 0.000933 = [\ce{H^+}]$.
So, the $\mathrm{pH} = -\lg([\ce{H^+}]) = -\lg(0.000933) = 3.03$
Looks pretty close to "B" to me!