Right choice is (c) 1.62
To explain I would say: Given, M\(_∞^2\) = 2, θ = 1.4 deg
For the supersonic flow, the linearized coefficient of pressure for small inclinations is given by:
Cp = \(\frac {2θ}{\sqrt {M_∞^2 – 1}}\)
Substituting the values, we get
Cp = \(\frac {2 × 1.4}{\sqrt {4 – 1}}\) = 1.62