The molecular formula of a compound with an empirical formula ${\mathit{CH}}_{2 }$and vapour density of 21 is:

Empirical formula weight (E.F.W) of ${\mathit{CH}}_2$
$\Rightarrow 12 + 2$  $\Rightarrow 14$ g
And the molecular weight of the compound is calculated by the formula,
$\Rightarrow$Molecular Weight = 2 $\times$ Vapour density
$\Rightarrow 2 \times 21$  $\Rightarrow 42$g
Now, we divide the molecular weight by the empirical formula weight, to obtain the subscript which will be multiplied by the empirical formula to obtain the molecular formula. So,
$\Rightarrow$n = M.V/E.F.W 
$\Rightarrow$ $\frac{42}{14}$
$\Rightarrow 3$
Hence molecular formula will be,
$\Rightarrow (C{H}_2{)}_n$ $\Rightarrow$ $(C{H}_2{)}_3$
$\Rightarrow C_3{H}_6$  

Hence, $C_3{H}_6$is the molecular formula of the compound.