The least number of years in which a sum of money on 19% p.a. compound interest will be more than double is

$A=P\left(1+\frac{R}{100}\right)^{T}$

$\Rightarrow 2 \mathrm{P}=\mathrm{P}\left(1+\frac{19}{100}\right)^{\mathrm{T}}$

$\Rightarrow 2=\left(\frac{119}{100}\right)^{\mathrm{T}}$

$\Rightarrow 2=(1.19)^{\mathrm{T}}$

If $\mathrm{T}=4$ years,

$(1.19)^{4}>2$