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RSMSSB JE Civil Diploma Sep 2016 Official Paper

Option 3 : 4.0 MPa

The maximum allowed shear given in IS code is based on the following formula:

\({\rm{\tau }} = 0.625\sqrt {{{\rm{f}}_{{\rm{ck}}}}} \)

Where, f_{ck} is the characteristic compressive strength

For M40: f_{ck} = 40 MPa

\({\rm{\tau }} = 0.625\sqrt {{{\rm{f}}_{{\rm{ck}}}}} = 0.625 \times \sqrt {40{\rm{\;}}} \) = 3.95 MPa **≈ 4 MPa.**