$(\,f_v\,)_{max} = \dfrac{VQ_{NA}}{Ib}$
Where
$(\,f_v\,)_{max} = 1.4 \, \text{ MPa}$
$Q_{NA} = 200(80)(140) + 100(80)(50)$
$Q_{NA} = 2\,640\,000 \, \text{ mm}^3$
$I = \dfrac{200(360^3)}{12} - \dfrac{120(200^3)}{12}$
$I = 697\,600\,000 \, \text{ mm}^4$
$b = 80 \, \text{ mm}$
Thus,
$1.4 = \dfrac{V(2\,640\,000)}{697\,600\,000(80)}$
$V = 29\,595.15 \, \text{ N}$
Spacing of bolts
$s = \dfrac{RI}{VQ_{flange}}$
$s = \dfrac{8(1000) \times 697\,600\,000}{29\,595.15 \, [ \, 200(80)(140) \, ]}$
$s = 84.18 \, \text{ mm}$ answer