Percentage

Number of girls that left the school = $120 \times \dfrac{5}{100} = 120 \times \dfrac{1}{20} = 6$

Remaining number of students = $120-6+57=171$

New percentage of boys = $\dfrac{57}{171} \times 100\% = \dfrac{1}{3} \times 100\% = \boxed{33\dfrac{1}{3}\%}$

Dumb problem. 57 is odd. The only odd option is D. It doesn't even need to be solved. !!

$5\%$ of the girls leave, so $95\%$ should be left which is $\dfrac{95}{100} × 120 = 114$ . So, total students = $114 + 57 = 171$ . So, percentage of boys is ${\dfrac{1 \cancel{57}}{3 \cancel{171}} × 100} = \color{#69047E}{\boxed{33 \frac{1}{3}}}$ .

Calculated exactly the part of the boys will be $33.\overline{3}\% = \frac{57}{0.95*120+57} = \frac{57}{171}$ ! 33.5% is located most nearby.