For neutralization, no. of gram equivalents of acid should be equal to no. of gram equivalents of base.
No. of gram equivalents of NaOH
$ = \frac{500 \times 0.2}{1000} = 0.1 $
Now, $ \frac{w}{eq. mass \, of\, acid} = 0.1$
eq. mass of acid = $ \frac{w}{0.1 } = \frac{6.3}{0.1 } = 63$