$0.2$ mole of an alkane on complete combustion gave $26.4$g of $C{O}_2$. The molecular weight of alkane is 

$ALKANE --------> C{O}_2$

$0.2$ mol     gives      $26.4$g

$1$mol        gives      $26.4/0.2$

$1$ mol        gives       $132$g

$1$ mol of $C{O}_2$  = $44$g

$3$mol of $C{O}_2$  = $132$g

It is concluding that alkane has three carbons and it is providing $3$ $C{O}_2$  

Then, the formula for alkane is $C_3{H}_8$ that is propane.

Molecular weight of propane = $44$

Therefore, Option (C) is correct.