Explanation:
The given data is as follows.
       [HCOOH] = 0.2 M,       [NaOH] = 2.0 M,
          V = 500 ml,   [Benzoic acid] = 0.2 M
First, we will calculate the number of moles of benzoic acid as follows.
    No. of moles of benzoic acid = Molarity × Volume
                          = 
                          = 0.095 mol
And, moles of NaOH present in the solution will be as follows.
     No. of moles of NaOH = Molarity × Volume
                           = 
                           = 0.05 mol
Hence, the ICE table for the chemical equation will be as follows.
          
Initial:        0.095           0.05            0             0 
Equlbm:  (0.095 - 0.05)  0            0.05
         pH = 
   
               = 
               = 4.245
For,   
          
Initial:       0.2x     2(0.5 - x)               0 
Equlbm:   0.2x - 2(0.5 - x)                 0             2(0.5 - x)
As, 
            pH = 
   
           4.245 = 3.75 + 
 
       
 = 0.5
     
 = 3.162
Now, 
         
 = 3.162
                x = 0.464 L
 Volume of NaOH = (0.5 - 0.464) L
                              = 0.036 L
                              = 36 ml               (as 1 L = 1000 mL)
And, volume of formic acid is 464 mL.