If a sum of money doubles itself at a fixed rate of compound interest per annum in $n$ years, find in how many years it will become four times.

Let the initial sum of money be $P$ and the rate of compound interest per annum be $R\%$.
According to the problem, the sum doubles itself in $n$ years:

$$2P=P{\left(1+\frac{R}{100}\right)}^n⟹2={\left(1+\frac{R}{100}\right)}^n\text{— (Equation 1)}$$

We want to find the time $T$ in which the sum becomes $4$ times its original value:

$$4P=P{\left(1+\frac{R}{100}\right)}^T⟹4={\left(1+\frac{R}{100}\right)}^T$$

We can express $4$ as $2^2$:

$$2^2={\left(1+\frac{R}{100}\right)}^T\text{— (Equation 2)}$$

Substitute the value of $2$ from Equation 1 into Equation 2:

$${\left[{\left(1+\frac{R}{100}\right)}^n\right]}^2={\left(1+\frac{R}{100}\right)}^T$$

$${\left(1+\frac{R}{100}\right)}^{2n}={\left(1+\frac{R}{100}\right)}^T$$

Comparing exponents on both sides:

$$T=2n\text{ years}$$