100 ~mL of 2 M of formic acid (p K_a=3.74) is neutralise by NaOH ,...

Sodium formate is present at the equivalence point. It is the salt of weak acid $+$ strong base. So, final solution will be basic in nature. As we know $\mathrm{pH}=7+\frac{\mathrm{p} K_a}{2}+\frac{\log C}{2}$ where, $C$ is concentration of salt. Total volume of solution $=100+100=200 \mathrm{ mL}$ Concentration of salt $(C)=2 \times \frac{100}{200}=1 \mathrm{M}\mathrm{pH}=7+\frac{3.74}{2}+\frac{\log [1]}{2}\mathrm{pH}=7+1.87+0\mathrm{pH}=8.87$