Explanation :
Given,$ \style{font-family:'Times New Roman'}{\begin{array}{l}{\mathrm V}_\mathrm u=200\;\mathrm{kN},\;{\mathrm\tau}_\mathrm c=0.62\;\mathrm{MPa},\\\end{array}} $
$ \style{font-family:'Times New Roman'}{\begin{array}{l}\;{\mathrm\tau}_\mathrm{cmax}=2.8\;\mathrm{MPa}\\\Rightarrow\;{\mathrm\tau}_\mathrm v=\frac{{\mathrm V}_\mathrm u}{{\mathrm b}_\mathrm d}=\frac{200\times10^3}{250\times350}=2.286\;\mathrm{MPa}\end{array}} $
As $ \style{font-family:'Times New Roman'}{\;{\mathrm\tau}_\mathrm v<\;{\mathrm\tau}_{\mathrm c\;\max}} $
and design shear froce $ \style{font-family:'Times New Roman'}{\left(\tau_v-\tau_c\right)\mathrm{bd}} $
$ \style{font-family:'Times New Roman'}{\begin{array}{l}=(2.286-0.62)\times250\times350\\=145.775\;\mathrm{kN}\end{array}} $
Spacing of shear reinforcement calculation :
$\style{font-family:'Times New Roman'}{\begin{array}{l}{\mathrm V}_\mathrm{us}=145775\\\;\;\;\;\;=0.87\times250\times2\times\frac{\mathrm\pi}4\times10^2\times\frac{350}{{\mathrm S}_\mathrm v}\\\Rightarrow{\mathrm S}_\mathrm v=82.03\;\mathrm{mm}\end{array}}$
Spacing for minimum shear reinforcement
$ \style{font-family:'Times New Roman'}{\begin{array}{l}\frac{{\mathrm A}_\mathrm{sv}}{{\mathrm{bS}}_\mathrm v}\geq\frac{0.4}{0.87{\mathrm f}_\mathrm y}\\\Rightarrow{\mathrm S}_\mathrm v\leq\frac{0.87{\mathrm f}_\mathrm y{\mathrm A}_\mathrm{sv}}{0.4\mathrm b}\\\Rightarrow{\mathrm S}_\mathrm v\leq341.65\;\mathrm{mm}\end{array}} $
Spacing should be minimum of
(i) 0.75 d = 26.25 cm
(ii) 8.2 cm
(iii) 34.16 cm
(iv) 30 cm
So spacing will be 8.2 cm