For the reaction
${C_2}{H_5}OH + HX\xrightarrow{{ZnC{l_2}}}{C_2}{H_5}X$
The order of reactivity is :
a.) HBr > HI > HCl
b.) HI > HCl > HBr
c.) HI > HBr > HCl
d.) HCl > HBr > HI

Hint: The reaction given is nucleophilic substitution reaction. The nucleophile is halide ion. The reactivity will thus base on the breaking of the H-X bond. Larger the size of the halide ion, easily it will be broken. So, greater will be its reactivity.

Complete answer:
The reaction given in question is nucleophilic substitution reaction. First, let us see what a nucleophilic substitution reaction is. These reactions involve a nucleophile which is an electron rich species that attacks on substrates to give product.
In the above reaction given to us, the alcohol is a substrate which is attacked by halide ions (nucleophile) to give alkyl halide product.
The formation of halide ions (${X^ - }$) depends on the strength of the H-X bond. If the H-X bond is strong, it will be difficult to break and if it is weak, it will be easy to break.
The Iodine is large in size and thus H-I bond will be easy to break. The iodide ion will be a strong nucleophile. So, the order of nucleophile will be -
Cl < Br < I
Thus, the order of reactivity is -
HI > HBr > HCl

So, the option (C) is the correct answer.

Note:
It must be noted that there is a large size difference between iodine and hydrogen. So, the bond is large. Thus, it is easy to break and will produce ions easily. The size difference between bromine and hydrogen is smaller as compared to difference between iodide and hydrogen. So, it will take more energy to break than HI. So, it will react slower than iodide.