$f_v = \dfrac{VQ}{Ib}$
Where
$b = x$
$Q = \frac{1}{2}xy\,(\frac{1}{2}h - \frac{2}{3}y) = \frac{1}{2}xy \left( \dfrac{3h - 4y}{6} \right)$
$Q = \frac{1}{12}xy(3h - 4y)$
$f_v = \dfrac{V \, [ \, \frac{1}{12}xy(3h - 4y) \,]}{Ix} = \dfrac{V}{12I}(3hy - 4y^2)$
$\dfrac{df_v}{dy} = \dfrac{V}{12I}(3h - 8y) = 0$
$3h = 8y$
$y = \frac{3}{8}h$
Location of maximum horizontal shearing stress:
$d = \frac{1}{2}h - y = \frac{1}{2}h - \frac{3}{8}h$
$d = \frac{1}{8}h$ answer