For a sum \(P\) at rate \(r\%\) per annum over 2 years:

Simple interest: \(SI = \dfrac{2Pr}{100}\).

Compound interest: \(CI = P\left[\left(1+\dfrac{r}{100}\right)^2 - 1\right] = P\left[\dfrac{2r}{100} + \dfrac{r^2}{10000}\right]\).

So \(\dfrac{CI}{SI} = \dfrac{\frac{2r}{100}+\frac{r^2}{10000}}{\frac{2r}{100}} = 1 + \dfrac{r}{200}\).

Given \(\dfrac{CI}{SI} = \dfrac{26}{25}\), so \(1+\dfrac{r}{200} = \dfrac{26}{25} \Rightarrow \dfrac{r}{200} = \dfrac{1}{25}\).

Solving, \(r = \dfrac{200}{25} = 8\). The rate of interest is 8% per annum.