Without any more information, it sounds like interest is compounded continuously, in which case the value of the investment A is given by
[tex]A=Pe^{rt}[/tex]
where P is the principal investment, r is the interest rate, and t is the number of years.
At the end of the second year (t = 2), the value is A = 60,500, and after the fourth year (t = 4), the value is A = 73,205. So solve the system
[tex]\begin{cases}60,500=Pe^{2r}\\73,205=Pe^{4r}\end{cases}[/tex]
for r. You can eliminate P by dividing
[tex]\dfrac{Pe^{4r}}{Pe^{2r}}=\dfrac{73,205}{60,500}\implies e^{2r}=\dfrac{121}{100}[/tex]
Take the logarithm (log here means natural log) of both sides to get
2r = log(121/100)
r = 1/2 log(121/100)
r ≈ 0.0953
So the interest rate is about 9.53%.
