The value of an investment in Canada Savings Bonds is modelled by $A(t) = A_0 e^{0.0255t}$... Rest of question below. The value of an investment in Canada Savings Bonds is modeled by $$A(t) = A_0 e^{0.0255t}$$, where A is the amount the investment is worth after $t$ years, and $A_0$ is the initial amount invested. At what rate, correct to 3 decimal places, is the investment growing at the time when its value has doubled?

I understand that we have to find the derivative but how can we find it if we don't know an initial amount?
The question asks for rate when the investments value has been doubled or when

$$A(t)=2A_0 = A_0 e^{0.0255t}$$
So the initial value cancels and you have:
$$
2=e^{0.0255t}\Rightarrow \ln(2)=0.0255t\Rightarrow t=\frac{\ln(2)}{0.0255}
$$
Now, taking the derivative:
$$A'(t)=0.0255A_0e^{0.0255t}\Rightarrow A'(\frac{\ln(2)}{0.0255})=0.0255A_0e^{0.0255(\frac{\ln(2)}{0.0255})}\\
=.0255A_0e^{\ln(2)}=2(.0255)A_0=.051A_0$$
Which does depend on the principal, as it should.
You have $A = 2A_0 \implies 2A_0 = A_0e^{0.0255t} \implies e^{0.0255t} = 2\implies 0.0255t = \ln 2 \implies t = \dfrac{\ln 2}{0.0255}$. To find the rate, we have: $A'(t) = 0.0255A_0e^{0.0255t} \implies A'\left(\frac{\ln 2}{0.0255}\right) = 0.0255A_0e^{0.0255\cdot \left(\frac{\ln 2}{0.0255}\right)} = 0.0255A_0e^{\ln 2} = 0.051A_0$.