I have asked OP to verify the solution concentration but didn't get the answer. Thus, I assume it is $8\% \ (w/w)$. Thus, if you assume $[\ce{HA}] = c$ then:
$$ c = 8\% \ (w/w) = \frac{\pu{8 g}\text{ of HA}}{\pu{100 g}\text{ of sol}} \times \frac{\pu{1.0 mol}\text{ of HA}}{\pu{60.05 g}\text{ of HA}} \times \frac{\pu{1.01 g}\text{ of sol}}{\pu{1.0 mL}\text{ of sol}} \times \frac{\pu{10^3 mL}\text{ of sol}}{\pu{1.0 L}\text{ of sol}}\\ = \pu{1.35 mol L-1}$$
Acetic acid ionization according to:
$$\ce{HA + H2O <=> H3O+ + A-}$$
If $\alpha$ amount of $[\ce{HA}]$ is ionized at equilibrium, concentrations at equilibrium would be $[\ce{HA}] = c - \alpha$, and $[\ce{A-}] = [\ce{H3O+}] = \alpha$. Thus:
$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]} = \frac{\alpha^2}{c- \alpha} \tag1$$
Since $c = \pu{1.35 mol L-1}$ and $K_\mathrm{a} = 1.77 \times 10^{-5}$, $c - \alpha \approx c$, the equation $(1)$ can be simplified here to get $\mathrm{pH}$. Take $\log$ on each side of the equation:
$$\alpha^2 = K_\mathrm{a} \times c \ \Rightarrow \ 2 \log \alpha = \log K_\mathrm{a} + \log c$$
Since $\alpha = [\ce{H3O+}]$:
$$2\times \mathrm{pH} = \mathrm{p}K_\mathrm{a} - \log c = 4.75 - 0.13$$
Thus, $\mathrm{pH} = 2.31$.
Late addition:
OP insists that his given answer for $\mathrm{pH}$ of the solution is $2.9$. Since I suspected OP's given data for the problem, I assumed the concentration of the acetic acid solution must be $0.8\% \ (w/w)$ instead of $8\% \ (w/w)$. When calculating with that value, you get $c = \pu{0.135 mol L-1}$. Hence,
$$\mathrm{pH} = \frac{1}{2}\left(\mathrm{p}K_\mathrm{a} - \log c \right) = \frac{1}{2}\left(4.75 + 0.87\right) = 2.81$$
- V is multiplied by 0,08 for 8% solution.
