$3^{\circ}>2^{\circ}>1^{\circ}$
The order of leaving group in $E 2$ elimination reaction is shown below :
$R - I > R - Br > R - Cl$
Therefore, the correct order for reaction with alcoholic $KOH$ is as follows :
$a > c > b > d$
${C_2}{H_5}OH + SOC{l_2}\xrightarrow{{{\text{Pyridine}}}}{C_2}{H_5}Cl + S{O_2} + HCl$
$(a)$ $\begin{array}{*{20}{c}}
{C{H_3} - CH - Br} \\
{|\,} \\
{\,\,\,\,\,\,{C_2}{H_5}}
\end{array}$
$(b)$ $\begin{array}{*{20}{c}}
{\begin{array}{*{20}{c}}
{\,\,Br} \\
|
\end{array}} \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,\,\,\,\,\,\,\,{C_2}{H_5}}
\end{array}$
$(c)$ $\begin{array}{*{20}{c}}
{C{H_3}{\mkern 1mu} - CH{\mkern 1mu} - {\mkern 1mu} {\mkern 1mu} C{H_2}Br} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{{C_2}{H_5}\,\,\,}
\end{array}$