$f_v = \dfrac{VQ}{Ib}$
Where
$V = 100(1000) = 100\,000 \, \text{ N}$
$I = \dfrac{120(240^3)}{12} - \dfrac{100(200^3)}{12} = 71\,573\,333.33 \, \text{ mm}^4$
$b = 20 \, \text{ mm}$
At the neutral axis
$Q_{NA} = 120(20)(110) + 20(100)(50)$
$Q_{NA} = 364\,000 \, \text{ mm}^3$
$(\,f_v\,)_{max} = \dfrac{100\,000(364\,000)}{71\,573\,333.33(20)}$
$(\,f_v\,)_{max} = 25.43 \, \text{ MPa}$ answer
At the junction of flange and web
$Q = 120(20)(110)$
$Q = 264\,000 \, \text{ mm}^3$
$(\,f_v\,)_{min} = \dfrac{100\,000(264\,000)}{71\,573\,333.33(20)}$
$(\,f_v\,)_{min} = 18.44 \, \text{ MPa}$
Average shear stress
$(\,f_v\,)_{ave} = (\,f_v\,)_{min} + \frac{2}{3} [\,(\,f_v\,)_{max} - (\,f_v\,)_{min}\,]$
$(\,f_v\,)_{ave} = 18.44 + \frac{2}{3} [\,25.43 - 18.44\,]$
$(\,f_v\,)_{ave} = 23.10 \, \text{ MPa}$
Shear force carried by web alone
Force = Stress × Area
$V_{web} = (\,f_v\,)_{ave} \, A_{web}$
$V_{web} = 23.10 (200 \times 20)$
$V_{web} = 92 400 \, \text{ N}$
$V_{web} = 92.4 \, \text{ kN}$
Percentage of shear force carried by web alone
$\% V_{web} = \dfrac{V_{web}}{V} \times 100\% = \dfrac{92.4}{100} \times 100\%$
$\% V_{web} = 92.4\%$ answer
